Math Adventures on CutTheKnot Math

Transcription

1 Mth Adventures on CutTheKnot Mth ROMANIAN MATHEMATICAL MAGAZINE Avilble online Founding Editor DANIEL SITARU ISSN-L

2 MATH ADVENTURES ON CutTheKnotMth By Alexnder Bogomolny nd Dniel Sitru 1

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4 Proposed by Dniel Sitru - Romni Hung Nguyen Viet - Hnoi - Vietnm Henry Ricrdo - USA Leonrd Giugiuc - Romni Abdilkdir Altints - Afyonkrshisr - Turkey Frncisco Jvier Grci Cpitn - Spin Anish Ry - Sntrgchi - Indi Mehmet Şhin - Ankr - Turkey

5 4 Solution by Dniel Sitru - Romni, Alexnder Bogomolny USA Nssim Nichols Tleb USA Henry Ricrdo - USA Soumitr Mndl - Indi, Amit Itgi Leonrd Giugiuc Romni, Rvi Prksh - Indi Kevin Soto Plcios - Peru, Soumv Chkrborty - Indi Abdul Aziz - Indonesi, Mrin Dincă - Romni Miquel Ocho Snchez - Peru, Andree Aquviv Seyrn Ibrhimov - Azerbidin, Chris Kyrizis - Greece Mygmrsuren Ydmsuren - Mongoli Mehmet Şhin - Turkey, George Apostolopoulos - Greece Redwne El Mellss - Morocco, Richdd Phuc - Vietnm Su Tny - Indi, Mrin Dniel Vsile - Romni Vsile Cîrtoje - Romni, Srinivs Vemuri Mrjn Milnovic - Serbi, Shivm Shrm - Indi Nirpd Pl - Indi, Nikolos Skoutris - Greece Nguyen Thnh Nho - Vietnm, Uche Eliezer Okeke - Nigeri Genin Tudose - Romni

6 101. Dn Sitru s Inequlity with Tngents II Prove tht in ny cute ABC, tn A tn B + 45 tn A tn B tn C Proposed by Dniel Sitru Solution 1by proposer. Let be x = cot A cot B; y = cot B cot C; z = cot C cot A. It follows tht x + y + z = 1. Further, 45 + = tn tn B = x + 1 y + 1 z = xy + yz + zx + 45xyz xyz 1 x + y + z + xy + yz + zx xyz + 46xyz xyz 1 x1 y1 z + 46xyz = xyz AM GM {}}{ = xyz x + 1 y + 1 z x + y + z xyz = xyz = cot A = tn A Solution by Kevin Soto Plcios. Set x = tn A > 0, y = tn B > 0, z = tn C > 0. Recollect tht x + y + z = xyz. The required inequlity is equivlent to which, in turn, is equivlent to xy + yz + zx + 45 x + y + z, x + y + z + xy + yz + zx 45. Since, by the rerrngement inequlity, x + y + z xy + yz + zx, suffice it to prove tht 5xy + yz + zx 45, i.e., tht xy + yz + zx 9, which follows from 1 xy + 1 yz + 1 zx = 1, by the Cuchy - Schwrz inequlity. 5 Solution by Soumv Chkrborty. tn A = tn A 6 tn A tn B becuse x + y + z xy + yz + zx. To continue, by the AM-GM inequlity, tn A tn B tn A tn B tn C = tn A = 9,

7 6 becuse tnx, being convex function on 0, π, A + B + C tn A tn Finlly. tn A 6 tn A tn B = tn A tn B + 5 tn A tn B tn A tn B Acknowledgment by Alexnder Bogomolny The problem from the Romnin Mthemticl Mgzine hs been kindly posted by Dn Sitru t the CutTheKnotMth pge. Dn lso hs communicted his solution Solution 1 in ltex file. Solution is by Kevin Soto Plcios; Solution is by Soumv Chkrborty. 10. A Cyclic Inequlity in Three Vribles XVII x y Prove tht, for x, y, z > 0, 5 x 9 y z x yz Proposed by Dniel Sitru Solution 1 by Leonrd Giugiuc. Let x y =, y z = b, z x = c. Then bc = 1 nd the required inequlity becomes c b c y z But, since + b + c, Hölder s inequlity gives 4. Also, c = b nd b, so tht 4 c. By the Rerrngement inequlity, 4 b. On the other hnd, 4 c = 6 b nd, ccording to n inequlity by Vsile Cîrtoje, 4 6 b. Finlly, 4, by the AM-GM inequlity. The required inequlity results s the product of 4 c, b, 4 [ 4 ] 6 b = 4 c, 4.

8 7 Solution by Nssim Nichols Tleb. 1 x y Prove tht, for x, y, z > 0, x y z 5 x y 9 x since y x y 1 x y + y z + z x x yz y z x y 4 xy z rerrngement inequlity x vrint of the rerrngement yz x y AM-GM since x y y z z 4 Finlly we need to prove tht 1 x x 4 y 8 + x y 6 z 4, x y y x y z z x = 1 x y z x y 7 z by some mysterious inequlity theorem I filed miserbly to find but tht we cn show vi clculus or some expnsion of the rerrngement inequlity. Postscript. Finnlly found the mysterious inequlity with help from Mestro Alexnder Bogomolny it necessittes minor chnge of vrible 1. If x, y, z re rel numbers, then x + y + z x y + y z + z x Vsile Cîrtoje, GM-B, 7-8, 199 Acknowledgment by Alexnder Bogomolny This problem hs been kindly posted t the CutTheKnotMth pge by Leo Giugiuc, long with his solution. The problem by Dniel Sitru hs been previously posted t the Romnin Mthemticl Mgzine. Solution is by N. N. Tleb. 10. A Cyclic Inequlity in Three Vribles XVIII Prove tht, for, b, c 0, 6 b 7 + b + b. Proposed by Dniel Sitru

9 8 Solution 1 by Kevin Soto Plcios. First note tht 1. + b + b 4 + b 0 b 0,., b, c > b b,., b, c R + b + c b + bc + c. It follows tht + b + b 7 + b 4 However, Thus, suffice it to prove tht 9 + b + c b + bc + c. 9 + b + c b + bc + c 7b + bc + c. 7b + bc + c b + bc + c 6. But, by the Cuchy - Schwrz inequlity, 7b + bc + c b + bc + c 8 b + bc + c 6. Solution by Soumitr Mndl. We know tht + b + b 4 + b + b b. Similrly, b + bc + c 4 b + c nd c + c + 4 c +, implying 7 + b + b 7 + b b 7 b b, 4 81 where we used the following inequlities: b 8 b,. + b + c b + bc + c, b+bc+c. b+ bc+ c. Solution by Dniel Sitru. Lemm 1 For x, y, z > 0, 1 xy x + xy + y Indeed, from Hölder s inequlity: x + xy + y = xy + y + x y + yz + z x + z + xz xy y x + y yz z + x z xz = xy + yz + zx.

10 9 Lemm For x, y 0, Indeed, nd, finlly, or, Similrly, xy 7 x xy + y. x y 0 x y 0 x 4xy + y 0 x xy + y x + xy + y. x xy + y 1 x + xy + y y yz + z 1 y + yz + z nd z zx + x 1 z + zx + x By multiplying the lst three reltionships, x + xy + y 7 x xy + y. Multiplying by 1 nd, 6 xy 7 x + xy + y x xy + y Thus, = 7 x + y x y = 7 x 4 + y 4 + x y. 4 6 xy = 7 x 4 + y 4 + x y Setting now in 4 x = ; y = b; z = c, we get 6 b 7 + b + b The equlity holds if = b = c. Solution 4 by Soumv Chkrborty. Cse 1: Exctly one of, b, c = 0 With = 0, the given inequlity reduces to 7b c b + bc + c b c, which is obvious. Cses b = 0 nd c = 0 re hndled similrly. Cse : At lest two of 0 0. Cse :, b, c > 0

11 10 Using Wu s inequlity, RHS 7b + bc + c. Thus, suffice it to prove tht 7b + bc + c b + bc + c 6 b + bc + c b + bc + c x x, where x = b, y = bc, z = c. The ltter is true by Hölder s inequlity. Solution 5 by Nssim Nichols Tleb. 1 We hve, by the Power Men Inequlity, b 1 1 b. Hence 6 LHS = b 1 b nd we need to prove tht b RHS, or 4 bc 4 b + b 4 + which is true by rerrngement. b c + b c + b c 0, Acknowledgment by Alexnder Bogomolny This problem hs been kindly posted t the CutTheKnotMth pge by Dniel Sitru nd then gin by Kevin Soto Plcios, long with his solution Solution 1. Solution is by Soumitr Mndl; Solution is by Dniel Sitru; Solution 4 is by Soumv Chkrborty; Solution 5 is by Nssim Nichols Tleb. The problem is by Dn Sitru nd hs been previously published t the Romnin Mthemticl Mgzine A Cyclic Inequlity in Three Vribles XXI Prove tht, for, b, c > 0, bc 7 7 Solution 1 by Leonrd Giugiuc. b 7 b+b +5b+b b + b + 5b + b. Proposed by Dniel Sitru Note tht. Indeed, tht is equivlent to b 7 b + 7b 0, which is true, with equlity for = b. Tking the product of three such inequlities we obtin the required one. Equlity when = b = c.

12 11 Solution Kevin Soto Plcios. Observe tht b + b = 4 b b b, nd + 5b + b = 4 b b b. Also, for, b, c > 0, + bb + cc + 8bc. Combining tht b + b 1 + 5b + b 4 + b 7 + b = 1 56 bc + bb + cc Solution by Soumv Chkrborty. b + b = 4 b b b. Similrly, b bc + c = Suffice it to prove tht which is the sme s Suffice it to prove tht which is true. b + c 4 nd c c + c +. Thus, 4 b + b + b b + c c b + b 8bc = + bb + cc + 8 bc + b bc bb + cc + + 5b + b 7 + b + 5b + b 7+b + 5b + b, which is equivlent to 7 + b + b 4 + 5b + b b 0, Solution 4 by Soumitr Mndl. The required inequlity is equivlent to 7 b + b bc + 5b + b. Now, 7 b + b b + 5b + b

13 1 = 7 + b 15b + b + b = b 7 + 7b b 0, so 7 b + b b + 5b + b. Rotting, b, c nd tking the product yields the required inequlity. Solution 5 by Rvi Prksh. Consider 7 b + b b + 5b + b = 7 b + b b b + b 6 b = 7 b + b 7b b + b + 6b b + b 6 b Hence, b+b +5b+b = 7 b + b b + b b + 6b b + b b = b [ b + b b ] 0 b 7. Rotting, b, c nd tking the product yields the required inequlity. Acknowledgment by Alexnder Bogomolny This problem form the Romnin Mthemticl Mgzine, hs been kindly posted t the CutTheKnotMth pge by Dniel Sitru. Solution 1 is by Leo Giugiuc; Solution is by Kevin Soto Plcios; Solution by Soumv Chkrborty; Solution 4 by Soumitr Mndl; Solution 5 by Rvi Prksh A Cyclic Inequlity in Three Vribles XXIII Prove tht, for, b, c > 0, + b + c 8bc + b + c + + b c Solution 1 by Abdul Aziz. Proposed by Dniel Sitru b bc 0 b + b c b c. Similrly, b c + c bc nd c + b bc. Adding up, b + b c + c bc + b + c. It s not hrd to see tht this inequlity is equivlent to the originl one just crry out the implied rithmetic nd cncel similr terms. Solution by Alexnder Bogomolny. We ll strt with b + b c + c bc + b + c. This is equivlent to b c + bc + c b + b + c. Assume, WLOG, tht b c. Then b c bc nd 1 c 1 b 1. By the Rerrngement inequlity, it then follows tht b c + bc + c b = b 1 c + bc 1 + c 1 b

14 1 = b 1 c + c 1 b + bc 1 b 1 b + c 1 + bc 1 c = + b + c. Solution by Kevin Soto Plcios, Soumv Chkrborty. In simplifiction of the bove rgument, observe tht x + y + z xy + yz + zx nd let x = bc, y = c, z = b to obtin b + b c + c bc + b + c. Solution 4 by Amit Itgi. Another wy of looking t b + b c + c bc + b + c: This is true due to Muirhed s theorem becuse the triple,, 0 mjorizes, 1, 1. Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted the problem from the Romnin Mthemticl Mgzine t the CutTheKnotMth pge nd lter dded the solution Solution 1 by Abdul Aziz. Solution hs been posted independently by Kevin Soto Plcios nd Soumv Chkrborty; Solution 4 is by Amit Itgi A Cyclic Inequlity in Three Vribles XXIV Lemm Prove tht, for, b, c > 0, b b b 4 bc + b + c. For x > 0, 1 + x 1 + x 1 Proof by lgebr by Dniel Sitru. x 1 0 Proposed by Dniel Sitru x x x + x x 1 + x 1 + x 1 + x 1 + x 1.

15 14 Proof by clculus by Alexnder Bogomony. If x = 1+x 1+x, f x = x +x 1 1+x, with the only positive zero x = 1, where f 1 = 1. Furthere, f x = 4 1+x 0, implying fx 1 Solution by Dniel Sitru. Be Lemm, 1 + b b 1 nd b b so tht, given the sequence,, 0 mjorizes, 1, 1, b b 4 1 b nd, subsequently, b b b 4 1 b b Muirhed {}}{ bc = 4 bc + b + c Inequlity is ttined for = b = c = 1. Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted the problem from the Romnin Mthemticl Mgzine t the CutTheKnotMth pge nd lter communicted the bove proof in LTeX file A Prtly Cyclic Inequlity in Four Vribles Prove tht, for x y z t, xe x x + y + e x+y+ + z + t e z+t Proposed by Dniel Sitru Solution by Dniel Sitru. Let fx = xe x, f x = x + e x. Thus for x, f is convex so tht, by Jensen s inequlity, fz + ft + f z + t f. i.e., z + t 1 fz + ft + f f For x, fx is concve. x + y = + x + y +. Thus, by Krmt s inequlity, fx + fy + f f + fx + y +

16 15 Adding up 1 nd gives z + t fz + ft + f + fx + fy f + fx + y + + f, i.e., ft fx + y + + f z+t or more explicitly, ze z + te t + xe x + ye y x + y + e x+y+ + z + t nd, finlly, e z+t. ze z + te t + xe x + ye y x + y + e x+y+ + z + t e z+t. Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted the problem from the Romnin Mthemticl Mgzine t the CutTheKnotMth pge nd lter communicted the bove solution in LTeX file A Problem From Mongolin Olympid for Grde 11 Prove tht, for, b, c > 0, subject to + b + c =, b b + c + b b + c + c c b + Solution by Dniel Sitru. By the AM-GM inequlity, + b = + b + c + b = + b + c + b + b 5 5 b 8 c 1 5, b + 1 c nd similr for the other two frctions, so tht, by the Rerrngement inequlity, + b 5 5 b 8 c 1 5 = 1 1 = 1 = 1 1 5b 8 c 1 5 5b 8 b Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted t the CutTheKnotMth pge the bove problem from the Mongolin Mthemticl Olympid, Grde An Aic Inequlity in Three Vribles Assuming, b, c > 0, prove tht bc + b c + c b + b + c + b + bc + c bb c. Proposed by Dniel Sitru

17 16 Solution by Dniel Sitru. b + b = b = 4 + b 4 + c 4 + b + b c + c b c b c bc + b + c = 4 bc + b c + b 4 b c + c +c 4 c b + b = bc It follows tht bc + b c + c b + b + c + b + bc + c = b + b + b = b Suffice it to prove tht b bb c + b + c b bc c b c b + bc + 4b + c 4b 4bc + c 0 b + c 0 Equlity is ttined for b = +c. Remrk by Alexnder Bogomolny It my be worth noting tht the inequlity t hnd is not ic, nd, in this sense, the ppernce of the ic sums my led the reder to think otherwise. The fct is tht the three vribles do not occur in the inequlity in symmetric mnner, so tht n rgument tht relies on th WLOG resoning my not be vlid. For exmple, if we ssume pprently WLOG tht c b, then the proof becomes immedite becuse bc + b c + c b + b + c 0 bb c. + b + bc + c However, such proof is fulty. Acknowledgment by Alexnder Bogomolny The problem bove hs been posted on the CutTheKnotMth pge nd the solution bove communicted to me by Dniel Sitru. Originlly, the problem hs been published t the Romnin Mthemticl Mgzine.

18 An Identity with Inrdius nd Circumrdii If I is the incenter of ABC, with r, R the inrdius nd circumrdius, nd R, R b, R c the circumrdii of tringles IBC, ICA, IAB. Prove tht R R b R c = R r Proposed by Mehmet Şhin Solution 1 by Dniel Sitru. In every tringle with side lengths, b, c re S nd circumrdius R, bc = 4RS. So, in IAB, r r AI BI AB c sin R c = = A sin B r 4[ IAB] 4 rc = sin A sin B etc. So tht r R = 8 r sin A = 8 Solution by Mrin Dincă. s bs c bc = r b c s 8S 4 = 16S r s 8S 4 = 16r s 8S = R r BC = R sin BIC = R sin π B + C = R sin B + C On the other hnd, BC = R sin A = R sina + B = 4R sin B + C cos B + C It follows tht R = R sin A, etc. We conclude tht R R b R c = 8R sin A sin B sin C = R r.

19 18 Solution by Miquel Ocho Snchez. If O is the circumcenter of IBC then O B = O I = O C = R. In AO B, O B = R sin α, implying R = R sin A. Similrly, R b = R sin B nd R c = R sin C. R R b R c = R 4R sin A sin B sin C = R r. Hence, R R b R c = R r. Acknowledgment by Alexnder Bogomolny The problem by Mehmet Şhin Turkey hs been posted to the site of the Romnin Mthemticl Mgzine nd communicted to me by Dniel Sitru. Solution 1 is by Dniel Sitru Romni; Solution is by Mrin Dincă Romni; Solution is by Miguel Ocho Snchez Peru An Inequlity In Tringle nd Without II In ABC,, b, c re the side lengths nd R, r re the circumrdius nd inrdius, respectively. Prove tht: c b b c c + c b + b 6r R Proposed by Dniel Sitru Remrk by Alexnder Bogomolny All solvers hve observed tht, due to Euler s inequlity R r, 6r R, nd went to prove c + c b b c c b + b Thus, this is the inequlity tht is proved below without dditionl comments. All solutions employ the AM-GM inequlity.

20 19 Solution 1 by Seyrn Ibrhimov. Set c = x, b c = y, b = z. The required inequlity becomes x + y + z + 1 x + 1 y + 1 z. We hve x + y + z x + y + 1 x z 6 + y z x y + 1 z z 6 xy =. x yz Solution by Dniel Sitru. First off, for ll x, y, z > 0, x + y + z = x + y + z + xy + yz + zx Thus, xy x + y + z + yz zx = x + y + z + 6 xyz 1 x + y + z x + y + z + 6 xyz. Set x = u + v w, y = u + w u, z = w + u v Then 1 gives u + v u + v w w + 6 u + v w u + v uv vw wu w uvw = u + v w + 6 u + vv + ww + u 8 = + 1 uvw It follows tht uv vw wu uvw + 10 = = 18. u + v v + w w + u w + +. u v With u =, v = b, w = c becomes the required inequlity. Solution by Soumv Chkrborty. b c + b + c + LHS c b x + yy + zz + x = 6, where x =, etc. xyz xy yz zx 6 = 6 8 = xyz

21 0 Acknowledgment by Alexnder Bogomolny This is problem form the Romnin Mthemtics Mgzine; it ws kindly communicted to me by Dniel Sitru, long with his solution Solution. Solution 1 is by Seyrn Ibrhimov; Solution is by Soumv Chkrborty. 11. An Inequlity in Tringle, Mostly with the Medins Prove tht in ny ABC, + m b m + 5m b < 64 5m s +. where s = + b + c. Proposed by Dniel Sitru Proof 1 by Soumv Chkrborty. By the AM-GM inequlity, 5m + m b m + 5m b 8m + m b, so tht 5m + m b m + 5m b 16m + m b. Now, obviously, m < b + c, m b < c +, m c < + b, implying 16m + m b < 4 + b + c = 4s + c. It follows tht 5m + m b m + 5m b 4s + c. We similrly obtin 5m b + m c m b + 5m c < 4s + nd 5m c + m m c + 5m < 4s + b The product of the three gives the required inequlity. Proof by Soumitr Mndl. First off, s + = b + + c + > c + b bc so tht + s > bc = 4 bc. On the other hnd, 5m + m b m + 5m b = 15m + m b + 4m m b c + + b + 4 c + b, becuse m = b + c, m b = + c b nd m m b c + b We need to prove tht c b + 17 b < 4s + c This is equivlent to 16c + b < + b + 16c + b, which is true becuse 4 c < + b nd b < + b. We only need to tke the product of this nd the two nlogous inequlities to obtin the result.

22 1 Acknowledgment by Alexnder Bogomolny Dniel Sitru hs kindly posted t the CutTheKnotMth pge the bove problem of his tht ws published in the Romnin Mthemticl Mgzine. Proof 1 is by Soumv Chkrborty; Proof is by Soumitr Mndl. 11. An Inequlity in Tringle, with Integrls [ For, b, c 0, π ], + b + c = π, prove tht 4 sin + π 0 cossin xdx π. Proposed by Dniel Sitru Solution by Dniel Sitru. By Kober s inequlity, cos x 1 x π such tht, sy cossin xdx 1 sin x dx = π π cos x 0 = + 1 sin π 1 = = 4 π sin Thus, ltogether,. It follows tht cossin x 1 sin x π 4 sin + π cossin xdx 0 4 sin + π 4 sin = π = π = + cos 1 π Refinement [by] Leonrd Giugiuc For, b, c, + b + c = π, prove tht 0, π π sin + 0 cossin xdx π It s known tht sin 4 nd cos t 1 t, t > 0. From the ltter, [ ] cossin x 1 sin x, for x 0, π. Suffice it to show tht 1 sin x 0 dx π 4. But 1 sin x dx = π sin x dx 0 0 = π + sin + π π = π 4.

23 Acknowledgment by Dniel Sitru This is Dniel Sitru s problem for the Romnin Mthemticl Mgzine. Solution is by Dniel Sitru. Leo Giugiuc cme up with n essentil refinement bove An Inequlity in Tringle, with Sides nd Medins Given ABC, with centroid G; side lengths, b, c nd the medins m, m b, m c. Prove tht 16 m + m 4 b 4+ b 4+ c 4 > 81. m c m c m m b m c Solution by Dniel Sitru. We hve self explntory sequence of steps: Proposed by Dniel Sitru GB + GC > BC GB + GC GB + GC > > GA GA GB + GC 4 4 > GA GA GB GA + GC 4 4 > GA m m b m + m 4 c m > m mb + m 4 c 81 4 > m m c 16 m mb 16 + m 4 c 4 > 81 m m m 16 m + m 4 b 4+ b 4+ c 4 > 81 m c m c m m b m c Acknowledgment by Alexnder Bogomolny The problem form the Romnin Mthemticl Mgzine hs been kindly posted by Dn Sitru t the CutTheKnotMth pge. Dn lso hs communicted his solution in ltex file.

24 115. An Inequlity in Tringle, with Sides nd Sums s 4s Solution by Dniel Sitru. Rewrite the inequlity s For ny ABC, prove tht + s b 4s b Using Bergstrom s inequlity, b + c + s c 4s c + b + c. b + c b + c b + c = [ Proposed by Dniel Sitru. ] b + c + b + c + = + = This is the required inequlity. Acknowledgment by Alexnder Bogomolny The problem from the Romnin Mthemticl Mgzine hs been kindly posted by Dniel Sitru t the CutTheKnotMth pge An Inequlity in Tringle, with Sines II For ny ABC, prove tht sin A sin A sin A sin B sin B sin 4 B Proposed by Dniel Sitru - Romni Solution 1 by Kevin Soto Plcios. In n tringle sin A, sin B, sin C > 0. In ddition, sin A sin B sin C 8. Thus pplying first Hölder s inequlity nd, subsequently, the AM-GM inequlity, sin A sin B sin A sin B sin A sin B 7 sin A sin B sin C 7 8 = 4. Equlity holds only for equilterl tringles. sin A sin B

25 4 Solution by Mygmrsuren Ydmsuren. In tringle sin A, sin B, sin C > 0. In ddition, 8 sin A sin B sin C = bc 8 R. Thus, 4 8R bc bc = 8R b b c c b b c c b b c c 8R b b b = = R b R sin A sin B R b R sin A sin B R b R sin A sin B Solution by Soumv Chkrborty. With the AM-GM inequlity, RHS 1 sin A sin A sin B sin B = 7 sin A 7 8 = 4 Solution 4 by Leonrd Giugiuc. Bsiclly, it s ll bout rerrngements: sin A sin B sin A sin B =, sin sin A sin B sin A sin A = 1 sin A, A+B+C = sin, by Jensen s inequlity, 180 = sin A sin B sin A sin A = 1 sin A = + cot A 4

26 5 Acknowledgment by Alexnder Bogomolny The problem form the Romnin Mthemticl Mgzine hs been kindly posted by Dniel Sitru t the CutTheKnotMth pge. Solution 1 is by Kevin Soto Plcios; Solution is by Mygmrsuren Ydmsuren; Solution is by Soumv Chkrborty; Solution 4 is by Leo Giugiuc An Inequlity Not in Tringle Prove tht for, b, c, d 0, + b b + b + c bc + c + d cd 6 + Solution by Dniel Sitru. + d. Proposed by Dniel Sitru With the reference to the bove digrm, OA =, OB = b, OC = c, OD = d. By the Lw of Cosines, AB = b b cos 45 = + b b, BC = b + c bc cos 0 = b + c bc CD = c + d cd cos 15 = c + d cd 6 +, AD + d, AB + BC + CD AD re the required inequlity follows. Acknowledgment by Alexnder Bogomolny This is problem nd solution from the Romnin Mthemticl Mgzine kindly communicted to me by Dniel Sitru.

27 An Inequlity with Ares, Norms, nd Complex Numbers Prove tht for, b, c, d R, with + b 0 nd c + d 0, one hs d bc + b c + d 4d bc 1 + b c + d Solution 1by Alexnder Bogomolny. Recollect tht * sin α = sin α 4 sin α Proposed by Dniel Sitru nd consider two points A =, b nd B = c, d in the plne with the origin O = 0, 0. The re [ ABO] cn be expressed in two wys, viz., [ ABO] = d bc, nd [ ABO] = + b c + d sin α, where α is the ngle t vertex O of the tringle. It follows tht d bc sin α + b c + d By noticing tht sin α 1 nd substituting into *, we obtin d bc 1 + b c + d 4 d bc + b c + d d bc + b c + d 4d bc = + b c + d which is the required inequlity. Solution by Rvi Prksh. Let z 1 = + ib, z = d + ic. Then z 1 z = d bc + ic + bd, d bc = 1 z 1z z 1 z, c + bd = 1 z 1z + z 1 z. Also, z 1 = + b nd z = c + d. Now, + b c + d 4d bc = z 1 z 1 z z 4 4 z 1z + z 1 z Set = [z 1z + z 1 z + z 1 z z 1 z z 1 z z 1 z ] = [z 1z + z 1 z z 1 z z 1 z ] Num = 1 z 1z + z 1 z [z 1z + z 1 z z 1 z z 1 z so tht = 1 [z 1z z 1 z ] Num 1 z 1z z 1 z 1 [ z 1z + z 1 z ] = z 1 z = + b c + d = Den, implying Num Den Num Den 1.

28 7 Solution by Rvi Prksh. Set z 1 = + ib = r 1 cos θ + i sin θ nd z = d + ic = r cos φ + i sin φ. z 1 z = d bc + ic + bd = r 1 r [cosθ + φ + i sinθ + φ]. Also, z 1 = r 1 nd z = r. Now, d bc + b c + d 4d bc + b c + d = r 1r cosθ + φ[r 1r 4r 1r cos θ + φ] r 1 r r = cosθ + φ 4 cos θ + φ = cos θ + φ 1 Solution 4by Soumitr Mndl. Let + b c + d = x, x > 0. We need to prove tht d bcx 4d bc x 1 x x d bc + 4d bc 0 x + d bc d bc[x d bc ] 0 x + d bcx xd bc + d bc x + d bc[x 4xd bc + 4d bc ] d bc[x d bc ] 0 x + d bcx d bc 0, which is true, for x bc d since c + bd 0, for, b, c, d R. Solution 5 by Soumv Chkrborty. From c + bd + d bc = + b c + d, + b c + d 4d bc = [c + bd + d bc ] 4d bc ] 1 {}}{ = c + bd d bc = c + bd d bc

29 8 Cse 1: c + bd 0, d bc 0 From the digrm, c + bd = p cos θ nd d bc = p sin θ,, where p = + b c + d. It follows tht LHS = d bcp cos θ p sin θ p, using 1,, 4 {}}{ = d bc cos θ sin θ. p Now, ccording s d bc 0 or d bc 0, d bc = ± d bc, 5. In ny event, d bc p = sin θ. Such tht, using 4 nd 5, LHS = ± sin θ cos θ sin θ = ± sin θ 1 sin θ sin θ = ± sin θ 4 sin θ = ± sin θ. Since sin θ 1, LHS 1. Cse : d bc = 0 Then c + bd 0 nd LHS = 0 1 Cse : c + bd = 0 d bc Then d bc = 0 nd LHS = = ±1 1 d bc Acknowledgment by Alexnder Bogomolny This is Problem SP060 from the 017 Spring issue of the Romnin Mthemticl Mgzine, proposed by Dniel Sitru. Solution nd re by Rvi Prksh, Solution 4 is by Soumitr Mndl; Solution 5 is by Soumv Chkrborty ll Indi.

30 119. An Inequlity with Circumrdii And Distnces to the Vertices Let M be n interior point of ABC. Denote by R, R b nd R c the circumrdii of the tringles MBC, MCA nd MAB, respectively. 9 MB MC R Solution 1by proposers. Prove tht MC MA MA MB + + MA + MB + MC R b R c Proposed by Leonrd Giugiuc, Abdilkdir Altints Denote MA = x, MB = y, MC = z, BMC = α, AMC = β, AMB = γ Clerly, 0 < α, β, γ < π nd α + β + γ = π. From the Lw of Cosines in MBC, BC = y yz cos α + z. Byt the y yz cos α + z Lw of Sines in MBC, R =. As y + z yz, sin α y yz cos α + z R = yz yz cos α sin α sin α = yz sin α yz sin α = cos α 1 such tht cos α.we hve R yz Similrly, MB MC R = yz yz cos α = yz cos α. R yz MC MA R b zx cos β

31 0 Thus, MB MC R MC MA + + R b Suffice it to show tht MA MB R c MA MB R c xy cos γ yz cos α + zx cos β + xy cos γ yz cos α + zx cos β + xy cos γ x + y + z But, ccording to the fmous Wolstenholme s inequlity, For rel x, y, z nd α + β + γ = π, yz cos α + zx cos β + xy cos γ x + y + z Replcing here x, y, z with x, y, z completes the proof. Solution by Dniel Sitru. Bsed on the following configurtion nd the formul bc = 4RS, MB MC MB MC = R 1 MB MC MC 4[ MBC] such tht, by similrity, MB MC MC MA MB MB + + R R b R c by the Erdös - Mordell inequlity. = 4 [ MBC] BC = MD, = MD + ME + MF MA + MB + MC Aknowledgment by Alexnder Bogomolny The problem due to Leo Giugiuc nd Kdir Altints hs been posted by Leo Giugiuc t the CutTheKnotMth fcebook pge, with their solution Solution 1 communicted privtely. Solution is by Dniel Sitru.

32 1 On the internl inequlity, 10. An inequlity with Cosines nd Sine Solution 1by Leonrd [ Giugiuc. 0, π Prove tht in cute ABC cos A + 4 cos B + 4 sin C 9 cos π + B C Proposed by Dniel Sitru ], function cos x is concve, such tht, by Jensen s cos A + cos B cos A + B = sin C Hence suffice it to prove tht cos B + cos A+B cos π+b C. We gin pply Jensen s inequlity: cos B + cos A + B = cos A + cos A + B B + A+B + A+B cos + cos A + B = cos A + B nd note tht, since A + B + C = π, A+B = π+b C Solution by Alexnder Bogomolny. π Observe tht sin α = cos α nd mke use of Jensen s inequlity s bove: cos A + 4 cos B + 4 sin C = cos A + 4 cos B + 4 cos π C 9 cos A + 4B + 4 π = 9 cos C π + B C 9 = 9 cos π + A + 4B C 9 = 9 cos π + B C. Aknowledgment by Alexnder Bogomolny The problem from the Romnin Mthemticl Mgzine hs been posted by Dn Sitru t the CutTheKnotMth pge, nd commented on by Leo Giugiuc with his solution Solution 1. Solution my seem s slight modifiction of Solution An Inequlity with Cyclic Sums And Products Prove tht if, b, c, d, e > 0 re pirwise distinct, then + b + c + d + e < b + c + d + e b c d e 104bcde Proposed by Dniel Sitru

33 Solution by Dniel Sitru. Let s split the left hnd side into prtil frctions: Then x cyc x k x k x = A = A + x + Bx x b This shows tht = B, C, D nd E. It follows tht A x = B x b + + b c d e. Cx x c C x c + + D x d + Dx x d + E x e Ex. x e Similr expressions cn be found for x b c d e x = x x bx cx dx e. Replcing x with + b + c + d + e nd using the AM-GM inequlity, we get b + c + d + e + b + c + d + e = k k [ k k] < + b + c + d + e + b + c + d + e = [4 4 k k] 4 4 bcde Aknowledgment by Alexnder Bogomolny The problem hs been kindly posted by Dn Sitru t the CutTheKnotMth pge; his solution in ltex file cme vi emil. 1. An Inequlity with Inrdius nd Circumrdii If I is the incenter of ABC, with r, R the inrdius nd circumrdius, nd R, R b, R c the circumrdii of tringles IBC, ICA, IAB. R R + R b + R c + R b + R c 1 6r R b R c R c R R R b R Proposed by Dniel Sitru Solution by Dniel Sitru. In every tringle with the side lengths, b, c re S, nd circumrdius R, we hve bc = 4RS. So, in IAB, Therefore, R c = R c = AI BI AB 4[ IAB] r sin A sin B r sin A r = 4 r = sin A sin B, r nd, similrly, R b = nd R = r sin B Thus, using Bergstrom s inequlity, R = r 1 sin B sin C r sin C sin A sin B sin C r sin B sin C.

34 It follows tht 1 Further, implying r 1 9 sin A = r 7 = r = 6r. R = 6r. r sin B sin C R = R b R c sin A sin B sin C = sin A r = 1 1 cos A = r r = 1 cos A r = r 1 r 1 + r R = r 1 r 1 R = r 1 R, R = R r R b R c Rr Multiplying 1 nd, we get R R r R 6R R b R c Rr = 6 R r R 6r = 1 R Acknowledgment by Alexnder Bogomolny This is Dn Sitru s problem from the Romnin Mthemticl Mgzine. Solution is by Dn Sitru. 1. An Inequlity with Integrls nd Rdicls Let f : [0, 1] 0, be continuous function such tht fxdx 5 fxdx 7 fxdx 1 0 Solution 1by Chris Kyrizis. By the AM-GM inequlity, fxdx = fx 1 1 fx so tht fxdx fxdx + Also Proposed by Dniel Sitru 1 5 fxdx = 5 fx 1 1 fx = fx +, dx = 1 = fx + 4, 5

35 4 so tht Similrly, fxdx 5 1 By multiplying 1, ; : fxdx fxdx 7 fxdx fxdx fxdx dx = 1 dx = 1 7 fxdx 1 Solution by Amit Itgi. Function x n is convex for x 0, n n integer so tht, by the integrl form of Jensen s inequlity n 1 1 [ ] n fxdx n ndx 1 fx = fxdx = It follows tht ech of the three integrls on the left is not greter thn 1, nd so is their product. Solution by Nssim Nichols Tleb. By the L p - norm inequlity, in its generl form, for ny L 1 function f nd for 0 < p q, 1 b p 1 b q fx p dx fx q dx Here pply p = 1 n j, q = 1, fx 0, for ll n j > 1: 0 Thus, b b Solution 4 by Andre Aquviv. Using Hölder s inequlity, with p = n, q = 1 0 j fx 1 n j dx nj 1. fx 1 n j dx nj p 1 g h dx h p dx g q dx n ndx n 1 fx n n 1, 1 p + 1 q = 1, 1 q 1 n fxdx = n fx 1dx 0 1 n 0 n 1 n n 1 dx, h = n f, g 1 : 1 = fxdx 0 1 n = 1.

36 5 Aknowledgment by Alexnder Bogomolny The problem from the Romnin Mthemticl Mgzine hs been posted by Dn Sitru t the CutTheKnot Mth pge. Dn lter communicted by emil his solution on LTeX file. The solution hs been obtined independently by Chirs Kyrizis. Solution is by Amit Itgi, Solution is by N. N. Tleb; Solution 4 is by Andre Acquviv. 14. An Inequlity with Powers of Six Prove tht in ny ABC, 6 + b 6 + c 6 8r s 5 b bc + c, where = + b + c, the semiperimeter of ABC, r its inrdius. Proposed by Dniel Sitru Proof by Soumitr Mndl, Seyrn Ibrhimov. From b 0, b + b b. So, using Euler s inequlity R r, bc = 4RS nd S = rs, RHS = 8r s b bc + c 8r s 5 bc 8r s 6 = 8r s 6 = 6 = LHS. bc 4RS Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted t the CutTheKnotMth pge the bove problem from his book Mth Accent, with proof by Soumitr Mndl Indi; Seyrn Ibrhimov Azerbijn submitted the sme proof independently An Inequlity with Tngents nd Cotngents Prove tht in ABC the following reltion holds tn A tn B + cot A cot B Proposed by Dniel Sitru Solution 1 by Leonrd Giugiuc. Set tn A tn B = z, tn B tn C = x, tn C tn A = y. We hve to prove tht x + 1 y + 1 z x y z 7, subject to x + y + z = 1. Consider function f : 0, 1 R, defined by ft = ln t + 1 t. We find f t = t t +1 1 t < 0, f 0 = 1 t t t > 0, for t 0, 1, implying tht the function is convex nd decresing so tht, vi Jensen s inequlity, x + y + z 1 fx + fy + fz f = f

37 6 i.e. ln x ln y ln z + 1 ln = ln, x y z implying the required inequlity. Solution by Leonrd Giugiuc. With the sme chnge of vribles, the problem reduces to x + 1 y + 1 z x y z 7, subject to x + y + z = 1. With Hölder s inequlity, we get x + 1 y + 1 z xyz + x y z xyz By the AM-GM inequlity, xyz 1, nd, since the function ft = t + 1 t is decresing for t 0, 1, 1 xyz = xyz 7, which completes the proof. Acknowledgment by Alexnder Bogomolny I m grteful to Dn Sitru for communicting to me his problem nd its two solutions by Leo Giugiuc. 16. An Inequlity with Vriety of Circumrdii Let R, R b, R c be the circumrdius of BOC, AOC respectively, AOB, where O is the circumcenter of n cute ABC. Prove tht R R b + R b R c + R c R R. Proposed by Dniel Sitru

38 7 Solution 1 by Mehmet Şhin. It s well known tht R = R cos A nd cos A cos B cos C 1 8. Thus, R R R b R c = 8 cos A cos B cos C R, R + R b + R c R + R b + R c = R + R b + R c, R b R c R R + R b + R c R + R b + R c R R b R c = R, by the AM-GM inequlity R + R b + R c R s desired. Solution by Soumv Chkrborty. By the property of centrl nd inscribed ngels subtended by the sme rc, BOC = A, thus, in the bove digrm, x = A. In OBP, OP = R cos x, implying OP = R cos A. It follows tht OB OC R R R = = 4[ OBC] 4 1 = OP R cos A = R cos A Similrly, R b = R cos B nd R c = R cos C. Therefore LHS = R cos B cos A + cos C cos B + cos A cos C R 1, due to the AM-GM inequlity cos A cos B cos C R becuse cyc cos A 1 8 Solution by George Apostolopoulos. It is well known tht R = R cos A, R b = R cos B, R c = R cos C nd cos A cos B cos C 1 8

39 8 So by AM-GM inequlity, we hve R R b + R b R c + R c R = R R b R c = R 1 = R. Equlity holds when ABC is equilterl. Solution 4 by Dniel Sitru. R R b R c R b R c R R 8 cos A cos B cos C [ BOC] = 1 OA OB sin BOC = 1 R sin A R = OB OC BC 4S[BOC] = R R 4 1 R sin A sin A = R sin A sin A cos A = R = cos A R AM GM + R b + R {}}{ c R R b R c R b R c R R b R c R = R R R b R c = 8 cos A cos B cos C R 8 1 = R. 8 Acknowledgment by Alexnder Bogomolny This is Dniel Sitru s problem from the Romnin Mthemticl Mgzine. Solution 1 by Mehmet Shin Turkey; Solution is by Soumv Chkrborty Indi; Solution is by George Apostolopoulos Greece; Solution 4 is by Dniel Sitru Romni who kindly provided tex file with ll the solutions. 17. An Inequlity with Just Two Vrible VII Prove tht, for x, y 0 x + y x xy + y x y xyx + y. Proposed by Dniel Sitru Solution 1by Kevin Soto Plcios. 1. x xy + y 1 x + y 1 4 x + y. Also x + y xy.. For x, y 0, x + y xy. The given inequlity is equivlent to Thus, x + y x xy + y x xy + y x y xyx + y x + y x xy + y 1 4 x + y4 x + y 8x y xy, x xy + y 1 8 x + y Multiplying the two gives the require inequlity.

40 9 Solution by Seyrn Ibrhimov. By Chebyshev s inequlity, By the AM-GM inequlity, Thus Becuse, x + y 1 x + yx + y x xy + y xy. x + y x xy + y xy 8 x + y x + y RHS x + y xy = 8xy xy Solution by Mygmrsuren Ydmsuren. Multiply the require inequlity by x + y which reduce it to: We hve xy x + y x + y 4 x y xyx + y x + y. 1 4 x + y 4 x + yx + y x + y x + y = x + y x + y x + y x + y xy xyx + y x + y. Solution 4 by Rvi Prksh. If x = 0 or y = 0, there is nothing to prove. Assume x, y > 0. Set x = r cos θ, y = r sin θ. The required inequlity reduces to r 11 cos θ + sin θ cos θ cos θ sin θ + sin θ r 11 cos θ sin θ 5 or, cos θ + sin θ 1 cos θ sin θ 4 cos θ sin θ 5. By the AM-GM inequlity, cos θ + sin θ cos θ sin θ. Also, since cos θ sin θ = sin θ < 1, 1 cos θ sin θ cos θ sin θ. In ddition, It follows tht 1 cos θ sin θ = 1 1 [1 cos θ sin θ ] = 1 [1 + cos θ sin θ ] LHS 8cos θ sin θ cos θ sin θ 1 8 [1 + cos θ sin θ ] = cos θ sin θ 5 [1 + cos θ sin θ ] cos θ sin θ 5 = RHS. Solution 5 by Alexnder Bogomolny. From x +y x +y, we hve x + y x + y. Also, x + y x y nd x xy + y xy. Multiplying ll three gives the required inequlity.

41 40 Solution 6 by Alexnder Bogomolny. The require inequlity is equivlent to x + x 4 xy 5 x + y x + y. Now, x + y x + y x + y xy x + y 1 x + y x + y 1 x y x + y = xyx + y The product of the three yields the required inequlity. Acknowledgment by Alexnder Bogomolny This problem from his book Algebric Phenomenon hs been kindly posted t the CutTheKnotMth pge by Dniel Sitru. Solution 1 is by Kevin Soto Plcios; Solution is by Seyrn Ibrhimov; Solution by Mygmrsuren Ydmsuren; Solution 4 by Rvi Prksh. 18. An Inequlity with Just Two Vrible VIII If, b [0, ] then b + + b + + b 1. Proposed by Dniel Sitru - Romni Solution 1 by Redwne El Mellss. Let f, b = b+ + b + + b. Then f0, 0 = 0, f > 0, 0 = 1; f0, b > 0 = b + b 1. Suppose, b > 0: 1. If b 1,. If b > 1, f, b + b b + = b + 1. f, b f0, b = = b + b b + + b b + + b + b b + = b + b 1 b b < 0, + b + implying tht f, b < f0, b = b = b 1. Finlly, f, b 1, with equlity if nd only if = 0 nd b =.

42 41 Solution by Richdd Phuc. We hve b + + b + + b 1 b b 8 + b + b 0 b b + b b b + + b + b 0 b b + b b b which is true becuse b + b b 0, b + b 4 + becuse, b [0, ]. Equlity holds if nd only if = 0 nd b =. Solution by Amit Itgi. b + + b + + b b b + 0, For fixed vlue of b, the right hnd side is liner function of with slope b+ b. The slope is non-negtive when b [0, 1] nd negtive when b 1, ]. In the regime where the slope is non-negtive, RHS ttins mximum vlue of 4 b+ + 4 t = the lrgerst vlue of. This function of b, in turn, ttins mximum vlue of 6 t b = 0. In the regime where the slope is negtive, RHS ttins mximum vlue of 4 + 4b t = 0 the smllest vlue of. Clerly, this function tkes mximum vlue of 1 when b = the lrgest vlue of b. Thus, LHS 1. Acknowledgment by Alexnder Bogomolny This problem from his book Algebric Phenomenon hs been kindly posted t the CutTheKnotMth pge by Dn Sitru. Solution 1 is by Redwne El Mellss; Solution is by Richdd Phuc; Solution is by Amit Itgi. 19. An Inequlity with Just Two Vribles III Prove tht, for positive, b, b + b + + b b + b + b 9. Proposed by Dniel Sitru Solution 1by Soumv Chkrborty. In right tringle with sides, b, + b. Let θ be the cute ngle opposite : = + b sin θ; : b = + b cos θ. Then

43 4 LHS = 1 tn θ + def {}}{ cot θ + cos θ + sec θ = = fθ Solving f 1 θ = cos θ 1 sin θ we get successively With t = sin θ > 0, Now, sin θ + sin θ cos θ = 0. sin θ cos θ cos θ sin θ sin θ cos θ 1 = 0, cos θ sin θ cos θ sin θ + θ sin cos θ = 0, sin θ cos θ sin θ = θ sin cos θ, sin θ = sin 5 θ, sin 4 θ 1 sin θ = 8 sin 10 θ. 8t 9t 4 + 1t 4 = 0, t 14t + t + t = 0, t = 1, sin θ =, θ = π 4.. f θ = sec θ tn θ + cot θ sin θ cos θ + sec θ + tn θ sec θ. f π 4 = 11 > 0, implying tht θ = π 4 is minimum nt tht f never ttends mximum on 0, π. Hence, π fθ f = = 9 Solution by Seyrn Ibrhimov. Note tht sin x + b cos x + b so tht sin x + b + b cos x + b 1, with equlity only when = + b sin x, b = + b cos x, x 0, π. The given inequlity is equivlent to where fx = 0 fx = 1 tn x + cot x + f x = cos x + cos x 9. 1 cos x + sin x sin x cos sin x = 0, x or, sin x 1cos x + 1 sin x 1 = 0 implies x = π π 4 f = 9 4.

44 4 Solution by Su Tny. Using the AM-GM inequlity, 9 b + b + b +b +b = [ b 1 + b [ 9 1 4] 1 b = + b 1 + b +b + b 8b becuse + b b 8b b + +b b + +b b + b ] 1 = 9 8b b 8b [ + b b 1 1 ] [ b = ] 1 = 9 1 = 9. Acknowledgment by Alexnder Bogomolny The problem bove hs been kindly posted to the CutTheKnotMth pge by Dn Sitru, long with severl solutions. Solutions 1 is by Soumv Chkrborty; Solution by Seyrn Ibrhimov; Solution is by Su Tny. 10. An Inequlity with Just Two Vribles V 1 Prove tht, for, b e, 1, 1 ln + 1 +b ln 1 b ln 1. b b Solution 1 by Dniel Sitru. 1 Define function f : e, 1 R by ln ln 1 x. Then f x = 1 x ln x f x = ln x x ln < 0, x e, 1. 1 Hence, f is concve on e., 1 By Jensen s inequlity, i.e., f b +b b f + b + bf +b ln + b b + fb +b. To continue, ln + b b b + b f + Proposed by Dniel Sitru + b f, b + b ln ln b ln ln 1, b

45 44 which is nd, finlly, 1 ln ln + 1 ln b 1 ln + 1 b ln 1 b +b ln 1 +b b ln 1 ln 1 1 +b b 1 ln + 1 +b ln 1 b ln b 1 b Solution by Nssim Nichols Tleb. Let x = 1 nd y = 1 b, x, y 1, e. We need to show tht ln x + y lnx x x+y lny y x+y Let y = x + ɛ, where 0 < ɛ < e x. We need to show tht ln x + ɛ lnx x x+ɛ lnx + ɛ x+ɛ x+ɛ. Expnding up to orders of ɛ, LHS = ln + ɛ x ɛ 8x ; RHS = ln x + ɛ x ɛ 8x ln x. So, since ln < 1, we hve LHS RHS. We cn refine further with the higher orders of Oɛ 4, by tking even orders. This is more engineering oriented but functionl pproch. Acknowledgment by Alexnder Bogomolny The problem bove hs been kindly posted to the CutTheKnotMth pge nd severl other forums in prticulr t the Romnin Mthemticl Mgzine by Dniel Sitru. After length of time tht it hd not gthered ny solution, Dniel hs communicted his solution by privte mil. Solution is by N.N. Tleb. 11. An Inequlity with One Tngent nd Six Sines tn A sin B + 5 sin C + Given n cute ABC. Prove tht tn B sin C + 5 sin A + tn C sin A + 5 sin B > 1 Proposed by Dniel Sitru Solution by Soumv Chkrborty. Note, tht, for α 0, π, tn α > α nd sin α < α. Thus, suffice it to prove tht A B + 5C + B C + 5A + C A + 5B 1 As in proof of Nesbitt s inequlity, the bove is equivlent to A AB + 5C + B BC + 5A + C CA + 5B 1

46 45 By Bergstrom s inequlity, A AB + 5C A 6 AB = A + AB 6 AB AB 6 AB 1 Refinement by Mrin Dincă Given n cute ABC. Prove tht tn A sin B + 5 sin C + tn B sin C + 5 sin A + tn C sin A + 5 sin B 1 Indeed, using the AM-GM inequlity, we obtin tn A sin B + 5 sin C tn A tn A tn B tn C sin B + 5 sin C = sin B + 5 sin C tn A = tn A tn A, implying However, tn A. To continue, sin B + 5 sin C Combining the ltest results, Equlity is ttined for A = B = C = π. sin B + 5 sin C = sin A. tn A sin B + 5 sin C = 1. Acknowledgment by Alexnder Bogomolny The problem from the Romnin Mthemticl Mgzine hs been posted by Dn Sitru t the CutTheKnotMth pge, Dn lter communicted solution by Soumv Chkrborty. Mrin Dinc cme up with the refinement of the originl inequlity. 1. An Inequlity with Sides, Altitudes, Angle Bisectors nd Medins Given ABC, with c b. Prove tht hb + h c + h hb + h c + h b m m b m c l l b l c + c b + c Proposed by Dniel Sitru Solution by Dniel Sitru. Note tht the condition c b implies cb c b 0. Using tht we prove tht b + c b + c b + b c + c. The two re equivlent s follows from the sequence below: b + c b + c b + b c + c

47 46 b c + c + b c + b + c b b c + cc bc 0 c b + c bc b 0 c bb c + c b 0 c b c b 0. To continue, since, sy h l m, b + c b + c b + b c + c h b h + h c h b + h h c Thus we hve nd lso h b + h c + h h b + h c + h l l b l c m m b m c h b m + h c m b + h m c b + c b + c h b Tking the product of the two gives hb s required. + h c + h b l + c b + c l b l c m + h c m b + h m c hb l + h c l b + h b l c + c b + c Aknowledgment by Alexnder Bogomolny The problem from the Romnin Mthemticl Mgzine hs been kindly posted by Dn Sitru t the CutTheKnotMth fcebook pge, Dn lter emiled me his solution in LtTex file. 1. An Inequlity with Sin, Cos, Tn, Cot, nd Some Prove tht, in cute ABC, the following inequlity holds: S sin A + cos A + tn A + cot A > 81πR 4 cos A, where S = [ ABC], the re nd R the circumrdius of ABC. Solution by Dniel Sitru. First for x we prove tht: 0, π Proposed by Dniel Sitru sin x + tn x > x Let be f : 0, π R, fx = sin x + tn x x. We hve f x = cos x + tn x 1 nd f c = sin x cos x cos x > 0. It follows tht f x > lim f x = 0 x 0 + such tht f x > 0, x 0, π, implying fx > lim fx = 0, x 0 +

48 so tht fx > 0, x 0, π. Hence, 1 sin x + tn x > x. Replce in 1 x with π x: π cos x + cot x > x 47 By dding 1 nd : sin x + tn x + cos x + cot x > π. For x = A; x = B; c = C in nd dding up: 4 sin A + cos A + tn A + cot A > π. By the AM-GM inequlity, + b + c +b+c bc such tht 8 7bc, i.e., 8s 8bc. Further 8s 7 4RS = 7 4Rrs, or, 7 Rr s nd, finlly 5 9Rr s. Now for few fcts, concerning the orthic tringle of ABC: the side length of the orthic tringle re: The inrdius: r = R cos A cos B cos C, The circumrdius: R = R, The semiperimeter: s = s R. Now, we pply 1 the orthic tringle: = cos A, b = b cos B, c = c cos c 9R r s, 9 R R cos A cos B cos C S R, cos A cos B cos C S 7R 4, 1 7R4 cos A S, S cos A 7R4, implying S 7R 4 cos A. Multiplying 4, 5: S sin A + cos A + tn A + cot A > 81πR 4 cos A. Acknowledgment by Alexnder Bogomolny I m grteful do Dn Sitru for posting this problem from his book Mth Accent t the CutTheKnotMth pge nd lter supplying LTeX file with its solution. 14. An Inequlity with Tngents nd Sides Prove tht in ny cute ABC the following reltionship holds: tn B + tn C + b tn C + tn A + c tn A + tn B sr. Proposed by Dniel Sitru

49 48 Solution by Dniel Sitru. tn C = BA HA, tn B = A C HA. tn B + tn C = BA + A C HA = HA HA = tn B + tn C S[BHC] = HA BC S[AHC] = S[AHB] = = 1 tn B + tn C b tn A + tn C c tn A + tn B S[ABC] = S[AHB] + S[AHC] + S[AHB] S = 1 tn B + tn C tn B + tn C = S = rs Euler {}}{ sr. Acknowledgment by Alexnder Bogomolny The problem from the Romnin Mthemticl Mgzine hs been posted by Dn Sitru t the CutTheKnotMth, Dn lter hs kindly communicted his solution in LTex file. 15. Another Integrl Inequlity from the RMM Ω = 1 1 Set x 6 + x 4 + rccosxdx, 1. Then Ω 19π 5 Proposed by Dniel Sitru

50 49 Proof by Rvi Prksh. For 1, where = 1 1 I 1 = Ω = 1 1 x 6 + x 4 + rccosxdx π x 6 + x 4 + rcsinx dx = π I 1 I, 1 1 x 6 + x 4 + dx = [ 7 x7 + ] 1 5 x5 + x 0 ] [ = [ ] 1 = 19 5, becuse 7 5. On the other hnd, I = 0, being n integrl of n odd function. So, finlly, Ω = π I 1 I π 19 5 = 19π 5. Acknowledgment by Alexnder Bogomolny The problem form the Romnin Mthemticl Mgzine hs been kindly posted t CutTheKnotMth pge by Dn Sitru, long with the solution by Rvi Prksh Indi. 16. Another Problem from the 016 Dnubius Contest Let, b, c > 0 stisfy b + bc + c =. Prove tht b c + 1. Proposed by Leonrd Giugiuc, Dniel Sitru Solution 1 by proposers. The inequlity is equivlent to b b + b c, or, 4 b + b c + c + b c let s denote bc = x, c = y, nd b = z. Then x, y, z > 0, x + y + z = nd xyz = b c. We hve to show tht x + y + z + xyz 4. We cn homogenize the inequlity: x + y + z x + y + z 4 x + y + z + xyz, reducing it to 5S s + xyz, where s = x + y + z nd s = xyx + y. From Schur s inequlity, S s + xyz 0 nd, from the well known inequlity u + v uvu + v, 4S s 0. Adding the two up gives the required inequlity.

51 50 Solution by Mrin Dniel Vsile is equivlent 1 which is + Bergstrom s inequlity: + b + c + + b + c + 6 = + b + c + b + bc + c + b + c + 6 Solution by Vsile Cîrtoje. nd, further, to + 1. To prove this we cn use + = + b + c + + b + c + 6 = 1. Let s denote bc = x, c = y, nd b = z. Then x, y, z > 0, x + y + z =. We hve to show tht x + y + z + xyz 4 Assuming x = min{x, y, z}, x 1, nd we hve x + y + z + xyz 4 = x + y + z + yzx 4 x + y + z y + z x 4 x + x + y + z 4 = x + x + 4 x 4 4 Solution 4 by Srinivs Vemuri. = 1 4 x 1 x + 0. Equlity occurs for = b = c = 1. c = b + b RHS LHS = c + 1 c + + b b + = b + + b b + + b + b b + Denomintors being positive, we ll focus on the numertors: + b + 4 b + + b + b + 4 b + + b + + b b + + b = + b b 4 + b 8b + b + b 4b b = + b b 6b b b b 4 b + 9b 8 = + b b 6b b b 4 b + 9b 6 + b = + b b 1b + bb 1 b b b bb 1 + bb 1 b b b

52 51 = bb 1 b + b + b = bb 1b 1b + + b Solution 5 by Amit Itgi. = bb 1 b + + b 0. We mke the sme substitution s in Solution 1 nd to obtin the problem of proving xyz + x + y + z 4 0, provided x + y + z = nd x, y, z > 0. Let x = k + m, y = k m, z = x y = k. The positivity of x, y, nd z implies > k > 0 nd k > m > k. We hve fk, m := k + mk m k + k + m + k m + k 4 = [5 kk 1 ] + [k 1m ]. Note: The first term the first squre brket is non negtive over the llowed rnge of k, becoming 0 t k = 1. The sign of the second term depends on k. Let us consider two cses: Cse 1: 1 > k > 0. In this cse, the second term is negtive n for fixed vlue of k is monotoniclly decresing function of m. Moreover, m < k implies f > 5 kk 1 + k 1k = 4k + 1 > 1 Cse : > k 1. In this cse, the second term is non negtive nd tkes vlue 0 when k = 1 or m = 0. Thus, f tkes minimum vlue of 0 when k = 1 nd m = 0. Solution 6 by Andre Aquviv. From b + bc + c =, + b + c, by the AM-GM inequlity. If we use D polr coordintes, then the problem reduces to finding the mximum of 1 ρ sin θ cos ϕ + 1 ρ sin θ sin ϕ + 1 ρ cos θ, where, s we just observed ρ. The bove function decreses s ρ grows o tht it chieves its mximum on the sphere ρ = - for the miniml vlue of ρ. Further, from b + bc + c = nd + b + c = it follows tht + b + c = 9, i.e. + b + c =, the tngent plne to the sphere + b + c t the point 1, 1, 1.

53 5 Illustrtion by Nssim Nichols Tleb Acknowledgment by Alexnder Bogomolny Leo Giugiuc hs kindly posted t the CutTheKnotMth pge the bove problem which he couthored with Dniel Sitru. The problem hs been included t 016 Dnubius contest. Solution 1 is by uthors; Solution is by Mrin Dniel Vsile; Solution is by Vsile Cîrtoje; Solution 4 is by Srinivs Vemuri; Solution 5 is by Amit Itgi; Solution 6 is by Andre Acquviv. The illustrtion is by N.N. Tleb. 17. Are Inequlity in Three Tringles Prove tht in three cute tringles A 1 B 1 C 1, A B C, A B C, S 1 + S + S < Proposed by Dniel Sitru

54 5 Solution 1 by Kevin Soto Plcios. First, recollect tht 4 S. By the Cuchy Schwrz inequlity, S 1 + S + S S 1 + S + S From here, S 1 + S + S 4 1 S 1 + S + S > S 1 + S + S Solution by Soumitr Mndl. By the Cuchy Schwrz inequlity, Sk k=1 S k. Also, bc 8s s bs c, where s = + b + c. It follows tht bc + b + c 4S. By the AM-GM inequlity, xy + yx + yz xyzx + y + z. Combining everything nd using the Rerrngement k=1 inequlity, 1 4 k b k + b k c k + c k k 1 4 k + b k + c k, k=1 implying k=1 k=1 S 1 + S + S 6 S k k + b k + c k < k=1 Solution by Soumv Chkrborty. By the Ccuchy Schwrz inequlity, S 1 + S + S 6S 1 + S + S. We shll prove tht in ny ABC, b 4 S. k=1 The ltter is equivlent to s + r4r + r 4 S. But s + r4r + r s r + rs, becuse s r nd 4R + r s. Thus, s + r4r + r 4 rs = 4 S, with conclusion tht 6S 1 + S + S 4 S 1 + S + S 1 b 1 + b + b 1 + +

55 54 Plese note tht we proved stronger inequlity thn required: S 1 + S + S < 1 b 1 + b + b. Solution 4 by Mygmrsuren Ydmsuren. S = sr nd s r, i.e., S s. With the AM-QM inequlity, S 1 4 [ Sk ] k= = < Remrk by Alexnder Bogomolny The required inequlity is rther wek. Ech of the vilble proofs estblished stronger inequlity: Prove tht in three cute tringles A 1 B 1 C 1, A B C, A B C, S 1 + S + S. Solution provided further refinement: Prove tht in three cute tringles A 1 B 1 C 1, A B C, A B C, 4 S 1 + S + S b. 1 b 1 + b + Equlity is chieved for three equl equilterl tringles. Acknowledgment by Alexnder Bogomolny The inequlity Romnin Mthemticl Mgzine hs been kindly posted t the CutTheKnotMth pge by Dniel Sitru. Solution 1 is by Kevin Soto Plcios Peru; Solution is by Soumitr Mndl Indi; Solution is by Soumv Chkrborty Indi; Solution 4 is by Mygmrsuren Ydmsuren Mongoli. 18. Cyclic Inequlity with Squre Roots Prove tht, for x, y, z 0, x + x z + y z xy xyz Proposed by Dniel Sitru

56 55 Solution 1 by Kevin Soto Plcios. Set x =, y = b, z = c. We choose, b, c 0. The given inequlity is equivlent to b + b c + c b z + b + c + c 4 + b 4 Let s prove c 4 + b 4 b bc + c. This is equivlent to c 4 + b 4 b + c + b c 4bcb + c nd, it turn, to b 4 + c 4 + b c + 4b c 4bcb + c = b + c 4bcb + c + 4b c = b c 4 0. So, we hve b bc + c c 4 + b 4 = b + b c + c bc + b + c b + b c + c bc + b + c + c 4 + b 4 Solution by Mygmrsuren Ydmsuren. xy = xy + zx = xy + z = x y + z = x x + xy + y = [ + ] x y + z yz x y + z + yz, by the Cuchy - Schwrz inequlity = x y + z x + xyz x = xyz x + y + z + x y + y z Acknowledgement by Alexnder Bogomolny Dn Sitru hs kindly posted this problem form the Romnin Mthemticl Mgzine, with two solutions, t the CutTheKnotMth pge. Solution 1 is by Kevin Soto Plcios; Solution is by Mymgsuren Ydmsuren. 19. Dn Sitru s Cyclic Inequlity in Three Vribles with Constrints Prove tht, for, b, c > 0, subject to + b + c = 6 + b + c, b b + c c b + c Proposed by Dniel Sitru

57 56 Solution 1 by Mrjn Milnovic. Since y = 1 x is convex function, b + c = + b = + b + c 1 Solution by Alexnder Bogomolny. By Bergstrom s inequlity, 1 LHS 9 + b + b + c + c +. LHS Now, using x x, + b + So to continue from 1, = b + b + c + c b + c = 1 + b + c Solution by Dniel Sitru. By Hölder s inequlity, 1 + b In other words, b 1 + b + b b + b = , + b implying, b + 6 = 7 7 = 1 This equivlent to the required inequlity.

58 57 Solution 4 by Shivm Shrm. Applying the AM-HM inequlity, b b + c + + b + c = b + c = 9 = b + c 1 + b + c Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted the CutTheKnotMth pge the bove problem from his book Algebric Phenomenon. Solution 1 is by Mki Milnovic; Solution is by Dn Sitru; Solution 4 is by Shivm Shrm Gireux s Theorem Theorem If continuous function of severl vribles is defined on hyperbrick nd is convex in ech of the vribles, it ttins its mximum t one of the corners. More formlly: Assume I k = [ k, b k ] R, k = 1, n nd f : I 1 I... I n R is continuous function convex seprtely in ech of the vribles in the domin of definition. Then it ttins its mximum t point C = c 1,..., c n where c k { k, b k }, k 1, n The sttement of the theorem is specifiction of theorem of Weierstrss the Extreme Vlues Theorem tht sttes tht continuous function defined on compct set ttins its extremes in set. Assume now tht the function is convex in ech vribles i.e., s function of one rgument, with other rguments fixed. A continuous function of one vrible, convex on closed intervl, ttins its mximum t one of the endpoints of the intervl. This mens tht the mximum of the given function is ttined t either, sy, I... I n or b 1 I... I n, which reduces the dimension of the serch for the mximum by 1. Doing this recursively proves the sttement. References: 1. Isrel Meireles Chrisostomo, Trigonometri Pur e Aplicções e um pouco lém: problems de Olimpíds, de Julho, 015 USA 1980 Prove tht, for, b, c [0, 1], b + c b c c b1 c 1. + b + 1 The function f, b, c = b+c is convex in ech of the three vribles, b, c so tht f tkes its mximum vlue in one of either vertices of the

59 58 cube 0 1, 0 b 1, 0 c 1. Since f, b, c tkes vlue 1 in ech of these points, the required inequlity is proven. References: 1. M. S. Klmkin, USA Mthemticl Olympids , MAA, 1988 Dn Sitru I Prove tht, for, b, c, d [0, ], bcd + 9b 1 + cd + 9c 1 + db + 9d 1 + bc + 9ebcd 8 + 9e 16. f : [0, ] 4 R, f, b, c, d = bcd + 9ebcd. f = bcd 9bcd 1 + cd 9cdb 1 + db 9dbc 1 + bc + 9bcdebcd, f = 18bc d 1 + cd + 18cd b 1 + db + 18db c 1 + bcd + 9b c d e bcd > 0. f strictly convex in vrible nd, similrly, in the rest of the vribles.f defined on compct set [0, ] 4, hence, by Gireux s theorem f ttins it mximum t the vertices of the hypercube [0, 1] 4. It is esy to check tht the mximum is ttined for f,,, = 4 the inequlity e16 = 8 + 9e 16, thus proving Dn Sitru II Prove tht, for x, y, z [0, 1] x y + z y z + x z + 1 x1 y1 z 1. x + y fx, y, z = f : [0, ] R, x y + z y z + x z + 1 x1 y1 z x + y f xx = We esily check tht y x + z z x + y > 0. fstrictly convex in vrible nd, similrly, in the rest of the vribles. f defined on compct set [0, ] 4, hence, by Gireux s theorem f ttins it mximum t the vertices of the hypercube [0, 1]. It is esy to check tht the mximum is ttined for f0, 0, 0 = 1, thus proving the inequlity.

60 59 Second proof Leo Giugiuc. x From x, y, z [0, 1] it follows tht y+z+016 x ; y z+x+016 y, z x+y+016 z. Thus suffice it to prove tht fx, y, z = x + y + z + 1 x1 y1 z 1. Let = 1 x, b = 1 y, c = 1 z. The inequlity to prove becomes b + 1 c + bc 1, or, bc +b+c, which is true becuse, for, b, c [0, 1], bc bc nd by the AM-GM inequlity. Third proof by Alexnder Bogomolny. Observe tht fx, y, z defined in the seconds proof is liner, hence convex, in ech of its rguments. Gireux s theorem pplies. f0, 0, 0 = 1, f0, 0, 1 = 1, f0, 1, 1 =, f1, 1, 1 = 1 = 1. Aknowledgment by Alexnder Bogomolny I m indebted to Dn Sitru for supplying the references nd the exmples Hung Viet s Inequlity III Prove tht, for ll rel, b, c 0, b b b 4 Solution by Kevin Soto Plcios. Observe tht by Hölder s inequlity 4 b b. Proposed by Hung Nguyen Viet b b, b The product of the bove two is exctly the required inequlity. Acknowledgment by Alexnder Bogomolny The inequlity by Nguyen Viet Hung hs been published t Spring issue of the Romnin Mthemticl Mgzine. This is Problem SP048. I reproduce here the chrming solution by Kevin Soto Plcios. Additionl solutions cn be found t the link.

61 Inequlity with Constrint XV nd XVI If, b, c re positive numbers such tht + b + c =, then b b c c + 4 > 5 b b c c > 5 Proposed by Henry Ricrdo Proposed by Dniel Sitru Problem 1, Solution 1 by Henry Ricrdo. WLOG,we my ssume tht b c. It follows tht b c nd b +4 1 c +4. Now the Rerrngement inequlity gives us b It cn be seen grphiclly nd proved with some tedious lgebr/nlysis tht the curve given by y = lies on or bove the tngent line to the curve t x x +4 x = 1, y = x , on the intervl 0,. Thus we hve b = = = 5 5 > 5. Problem 1, Solution by Alexnder Bogomolny. Using Bergström inequlity nd the obvious x + > x + 4, b b + c b + c > + = 9 9 = 1 Problem, Solution by Anish Ry. As bove, using Bergström inequlity, the obvious x + > x nd x + y + z x + y + z, + b + c b > b + c + 6 = 9 15 = 5. Acknowledgment by Alexnder Bogomolny The problem Problem 1 with solution Problem 1, Solution 1 hs been posted to the Romnin Mthemticl Mgzine by Henry Ricrdo. A somewht more complicted but similr problem hs been posted by Dn Sitru, with credits to Anish Ry.

62 14. Inequlity with Constrint from Dn Sitru s Mth Phenomenon If b c > 0 nd + b + c = 10, then b b + b b + c b Proposed by Dniel Sitru Solution 1 by Soumitr Mndl. Note tht + b + b + b 4 + b = + b = + b + c = + b + c + b + c + + b + c b + c + 0, + b 4 becuse c. This proves the right inequlity. The left inequlity is equivlent to 4 + b + c + b + b + b which, in turn is equivlent to + b + b + b + b + c b + bc + c + + b b + c + + c + c + c 0, + c or, b b bb c c c b b + c c + 0, which is true becuse b c. Solution by Alexnder Bogomolny. The left inequlity is equivlent to 4 + b + c + b + b + b = + b + b b + b + b = + b which cn be rewritten s b + b = 4 + b + c b + b b + c. b + b 61 Now note tht b + b + bc b + c b + bc b = b + c.

63 6 Also c c+ c is equivlent to c c + c, or c, which is true. Now, dding this to b + b + bc b + c b + c completes the proof of the left inequlity. For the right inequlity, observe tht, s we just showed, Thus suffice it to prove tht b + b b + c. + b + b 0. This is indeed so due to Bergström s inequlity: + b + b = + b + b + b + b + c + b + c + + b + c + b + c = + b + c = 0. Solution by Nssim Nichols Tleb. Let = +b+b +b. We reexpress: f = + b b = 0 b + b + b Let us further estblish tht from the ssumtions b c > 0 nd + b + c = 10, it is necessry tht nd c. We hve the right side inequlity: b + c < 0. We lso hve f 0, for b +b b from which b + b 1 + b + c = 5 For the left inequlity, we hve b + b + cb b + c + c c + b + bc b + c = b + c, so f 40 b c. The left side inequlity becomes b b c, i.e., + b + c 0 which is true. Acknowledgment by Alexnder Bogomolny The bove problem, originlly from his book Mth Phenomenon, hs been posted by Dn Sitru t the CutTheKnot Mth. Solution 1 is by Soumitr Mndl; Solution is by N. N. Tleb.

64 144. Problem 4 From the 016 Pn Africn Mth Olympid Let x, y, z be positive rel numbers such tht xyz = 1. Prove tht 1 x y Solution by Dniel Sitru. We ll first use the AM-GM inequlity: x y + 1 = x + y + x + xy + x + 1. The other two summnds re modified ppropritely. Introduce, b, c vi x = b, y = c b, z = c. Note tht xy = c. It follows tht 1 x y xy + x + 1 = 1 1 c + b + 1 = 1 + b + c = 1 + b + c + b + c = 1 Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted t the CutTheKnotMth pge the bove problem from the 4th Pn Africn Mthemticl Olympid, long with his solution. Dn hd lso remrked tht the problem ws creted in 006 by Cristinel Mortici from Romni Simple Inequlity with Vriety of Solutions Prove tht, for x, y, z > 1, ln x ln y ln z + ln y 18 ln z ln x lnxyz. Proposed by Dniel Sitru - Romni Solution 1by Kevin Soto Plcios. Let = ln x, b = ln y, c = ln z. The inequlity becomes bc + b b + c c + c c + b b bc b + c, or, equivlently, bc + b c + c + b + c 9. b This is seen to be true by first pplying the Cuchy - Schwrz nd then AM-GM inequlity: + b + c + b + c + b + c bc bc = + b + c bc 9.

65 64 Solution by Nirpd Pl. Let = ln x, b = ln y, c = ln z. The inequlity reduces to bc + b b + c c + c c + b b bc b + c, LHS = bc + b = + b b c b bc = 1 AM HM {}}{ 9 + b + c = 18 + b + c. Solution by Dniel Sitru. Let = ln x, b = ln y, c = ln z. The inequlity reduces to bc + b b + c c + c c + b b bc b + c, LHS = bc + b = c bc bc b = AM HM 1 {}}{ 9 bc + b + c 18 = + b + c Solution 4 Nikolos Skoutris. Let = ln x, b = ln y, c = ln z. The inequlity reduces to bc + b b + c c + c c + b b bc b + c LHS = bc + b = + b c bc b bc = 1 1 = + 1 b + 1 AM HM {}}{ c 6 Solution 5 by Nguyen Thn Nho. LHS = ln x ln y ln z + = 18 + b + c x, y, z > 1 ln x, ln y, ln z > 0 ln y ln z ln x b c AM GM {}}{ 1 ln x + 1 ln y + 1 ln z Cuchy Schwrz {}}{ ln x + ln y + ln z = 18 lnxyz.

66 65 Solution 6 by Soumv Chkrborty, Genin Tudose. Let = ln x, b = ln y, c = ln z. Then inequlity reduces to bc + b b + c c + c c + b b bc b + c or, equivlently, 1 9 bc + b + c, or, By the AM-GM inequlity 9. b c bc the product of which is 9bc, which is equivlent to the required inequlity. Solution 7 by Uche Eliezer Okeke. Let = ln x, b = ln y, c = ln z. The inequlity reduces to bc + b b + c c + c c + b b bc b + c, or, equivlently, 1 9 bc + b + c LHS = 1 bc AM GM {}}{ bc 7 1 = 18 Solution 8 by Seyrn Ibrhimov. Let = ln x, b = ln y, c = ln z. The inequlity reduces to bc + b b + c c + c c + b b bc b + c, LHS = Chebyshev {}}{ b + c bc bc Cuchy Schwrz {}}{ + b + c b + bc + c 9 + b + c becuse + b + c b + bc + c which follows from + b + c b + bc + c

67 66 Solution 9 by Mygmrsuren Ydmsuren. Let ln x = t 1, ln y = t, ln z = t. The inequlity reduces to LHS = t 1 = 1 t1 + t t t t t t 1 AM GM {}}{ Bergstrom 1 {}}{ t 9 t 1 Solution 10 by Alexnder Bogomolny. Let = ln x, b = ln y, c = ln z. The inequlity reduces to AM GM {}}{ 1 bc bc AM GM {}}{ 18 = t 1 + t + t 9 + b + c Acknowledgment by Alexnder Bogomolny Dn Sitru hs kindly posted the problem form the Romnin Mthemticl Mgzine t the CutTheKnotMth fcebook pge nd ltter commented with severl proofs. Solution 1 is by Kevin Soto Plcios; Solution is by Nirpd Pl; Solution is by Dn Sitru; Solution 4 is by Nikolos Skoutris; Solution 5 is by Nguyen Thnh Nho; Solution 6 is by Soumv Chkrborty, Genin Tudose cme up with the sme solution; Solution 7 is by Eliezer Okeke; Solution 8 is by Seyrn Ibrhimov; Solution 9 is by Mygmrsuren Ydmsuren; 146. Sitru Schweitzer Inequlity Below is slightly modified versions of Dn Sitru s sttement from his book Mth Phenomenon. The problem represents n integrl nlog nd generliztion of the well known Schweitzer s Inequlity derived next. Let f : [, b] [m, M], with m > 0, be n Riemnn integrble function such 1 tht is lso Riemnn integrble. Then fx b b 1 fxdx fx dx m + M b 4mM Remrkby Alexnder Bogomolny Note tht, by the Cuchy criterion for integrbility, if function f : [, b] [[m, M], ] with m > 0 is Riemnn integrble, then the function 1 f : [, b] 1 M, 1 m is lso Riemnn integrble. Solution 1 by Rvi Prksh. It follows from the premises tht f x + mm m + Mfx, or fx m M fx 0, implying fx + mm fx m + M,

68 67 so tht b fxdx + mm b dx m + M fx b dx = m + Mb Let J = mm b dx fx nd H = b fxdx. We hve J + H m + Mb, implying J + JH m + Mb J, i.e., JH 1 [ 1 ] 4 m + M b m + Mb J This is the required inequlity. Solution by Soumitr Mndl. We proceed s bove to obtin b fxdx + mm b 1 4 m + M b dx m + M fx By the AM-GM inequlity, b m + Mb mm Squring gives the required inequlity. Schweitzer s Inequlity b dx = m + Mb. b fxdx dx. fx For 0 < m < M, nd x k [m, M], for k 1, n, n n 1 m + M x k 4mM n. k=1 x k k=1 Proof of Schweitzer s Inequlity by Alexnder Bogomolny. Given sequence {x k } [m, M], we define function piece wise: set, for x [k, k + 1], fx = x k, k 1, n. Then, with = 1 nd b = n + 1, f : [, b] [m, M], nd it remins to observe tht, b b b = n n fxdx = dx fx = k=1 n k=1 x k 1 x k. Note tht two dditionl proofs pper elsewhere. Acknowledgment by Alexnder Bogomolny The problem hs been kindly posted by Dn Sitru t CutTheKnotMth fcebook pge. Solution 1 is by Rvi Prksh; Solution is by Soumitr Mndl.

69 The Beuty of Frctions On circle of rdius r, three points re chosen so tht the circle is divided into three rcs in the rtios u : v : w. At the division points, tngents re drwn to the circle: Prove tht the re of the tringle formed by the tngents equls S = ±r πu tn u + v + w tn πv u + v + w tn πw u + v + w. Proposed by Frncisco Jvier Grci Cpitn Solution by Dniel Sitru. Set AB = x + y, BC = y + z, AC = z + x. Scle the rc rtios to stisfy u + v + w = π. As we cn see, x = r tn u, y = r tn v, z = r tn w. Heron s formul reduces to S = xyzx + y + z, so we hve S = xyzx + y + z = r 4 tn u tn u = r tn u = ±r tn u = ±r πu tn u + v + w tn πv u + v + w tn πw u + v + w, where we pplied n identity vlid for α + β + γ = π. tn α + tn β + tn γ = tn α tn β tn γ Acknowledgment by Alexnder Bogomolny The problem by Frncisco Jvier Grcí Cpitán hs been kindly communicted to me by Dn Sitru, long with his solution.