Solutions of Inequalities problems (11/19/2008)

Transcription

1 Solutios of Iequalities problems (/9/8).[4-A] First solutio: (partly due to Ravi Vakil) Yes, it does follow. For i =,, let P i, Q i, R i be the vertices of T i opposide the sides of legth a i, b i, c i, respectively. We first check the case where a = a (or b = b or c = c, by the same argumet after relabelig). Imagie T as beig draw with the base Q R horizotal ad the poit P above the lie Q R. We may the positio T so that Q = Q, R = R, ad P lies above the lie Q R = Q R. The P also lies iside the regio bouded by the circles through P cetered at Q ad R. Sice Q ad R are acute, the part of this regio above the lie Q R lies withi T. I particular, the distace from P to the lie Q R is less tha or equal to the distace from P to the lie Q R ; hece A A. To deduce the geeral case, put r = max{a /a, b /b, c /c }. Let T 3 be the triagle with sides ra, rb, rc, which has area r A. Applyig the special case to T ad T 3, we deduce that A r A ; sice r by hypothesis, we have A A as desired. Remark: Aother geometric argumet i the case a = a is that sice agles Q ad R are acute, the perpedicular to Q R through P separates Q from R. If A > A, the P lies above the parallel to Q R through P ; if the it lies o or to the left of the vertical lie through P, we have c > c because the iequality holds for both horizotal ad vertical compoets (possibly with equality for oe, but ot both). Similarly, if P lies to the right of the vertical, the b > b. Secod solutio: (attributio ukow) Retai otatio as i the first paragraph of the first solutio. Sice the agle measures i ay triagle add up to π, some agle of T must have measure less tha or equal to its couterpart i T. Without loss of geerality assume that P P. Sice the latter is acute (because T is acute), we have si P si P. By the Law of Sies, A = b c si P b c si P = A. Remark: May other solutios are possible; for istace, oe uses Hero s formula for the area of a triagle i terms of its side legths..[4-b] First solutio: We have ( ) m + (m + ) m+ > m m m because the biomial expasio of (m + ) m+ icludes the term o the right as well as some others. Rearragig this iequality yields the claim. Remark: Oe ca also iterpret this argumet combiatorially. Suppose that we choose m + times (with replacemet) uiformly radomly from a set of m + balls, of which m are red ad are blue. The the probability of pickig each ball exactly oce is (m + )!/(m + ) m+. O the other had, if p is the probability of pickig exactly m red balls, the p < ad the probability of pickig each ball exactly oce is p(m m /m!)( /!). Secod solutio: (by David Savitt) Defie S k = {i/k : i =,...,k} ad rewrite the desired iequality as x S m x y S y > z S m+ z. To prove this, it suffices to check that if we sort the multiplicads o both sides ito icreasig order, the i-th term o the left side is greater tha or equal to the i-th term o the right side. (The equality is strict already for i =, so you do get a strict iequality above.)

2 Aother way to say this is that for ay i, the umber of factors o the left side which are less tha i/(m + ) is less tha i. But sice j/m < i/(m + ) is equivalet to j < im/(m + ), that umber is im + m + i m + im m + + Third solutio: Put f(x) = x(log(x + ) log x); the for x >, f (x) = log( + /x) x + f (x) = x(x + ). i m + = i. Hece f (x) < for all x; sice f (x) as x, we have f (x) > for x >, so f is strictly icreasig. Put g(m) = m log m log(m!); the g(m + ) g(m) = f(m), so g(m + ) g(m) icreases with m. By iductio, g(m + ) g(m) icreases with for ay positive iteger, so i particular g(m + ) g(m) > g() g() + f(m) g() sice g() =. Expoetiatig yields the desired iequality. 3.[3-A] First solutio: Assume without loss of geerality that a i +b i > for each i (otherwise both sides of the desired iequality are zero). The the AM-GM iequality gives ( ) / a a ( a + + (a + b ) (a + b ) a + b a a + b ad likewise with the roles of a ad b reversed. Addig these two iequalities ad clearig deomiators yields the desired result. Secod solutio: Write the desired iequality i the form (a + b ) (a + b ) [(a a ) / + (b b ) / ], expad both sides, ad compare the terms o both sides i which k of the terms are amog the a i. O the left, oe has the product of each k-elemet subset of {,...,}; o the right, oe has ( ) (a a ) k/ (b...b ) ( k)/, k which is precisely ( k) times the geometric mea of the terms o the left. Thus AM-GM shows that the terms uder cosideratio o the left exceed those o the right; addig these iequalities over all k yields the desired result. Third solutio: Sice both sides are cotiuous i each a i, it is sufficiet to prove the claim with a,..., a all positive (the geeral case follows by takig limits as some of the a i ted to zero). Put r i = b i /a i ; the the give iequality is equivalet to I terms of the fuctio ( + r ) / ( + r ) / + (r r ) /. f(x) = log( + e x ) ad the quatities s i = log r i, we ca rewrite the desired iequality as ( ) (f(s s + + s ) + + f(s )) f. ),

3 This will follow from Jese s iequality if we ca verify that f is a covex fuctio; it is eough to check that f (x) > for all x. I fact, f (x) = ex + e x = + e x is a icreasig fuctio of x, so f (x) > ad Jese s iequality thus yields the desired result. (As log as the a i are all positive, equality holds whe s = = s, i.e., whe the vectors (a,...,a ) ad (b,..., b ). Of course other equality cases crop up if some of the a i vaish, i.e., if a = b =.) Fourth solutio: We apply iductio o, the case = beig evidet. First we verify the auxiliary iequality (a + b )(c + d ) (ac + bd ) for a, b, c, d. The left side ca be writte as ( ) a c ( ) + b d ( ) + a c i d ( i) + i ( ) b c ( i) d (i ). i Applyig the weighted AM-GM iequality betwee matchig terms i the two sums yields ( ) (a + b )(c + d ) a c ( ) + b d ( ) + a i b i c ( )i d ( )( i), i provig the auxiliary iequality. Now give the auxiliary iequality ad the case of the desired iequality, we apply the auxiliary iequality with a = a /, b = b /, c = (a a ) /( ), d = (b... b ) /( ). The right side will be the -th power of the desired iequality. The left side comes out to (a + b )((a a ) /( ) + (b b ) /( ) ), ad by the iductio hypothesis, the secod factor is less tha (a +b ) (a +b ). This yields the desired result. Note: Equality holds if ad oly if a i = b i = for some i or if the vectors (a,...,a ) ad (b,..., b ) are proportioal. As poited out by Naoki Sato, the problem also appeared o the 99 Irish Mathematical Olympiad. It is also a special case of a classical iequality, kow as Hölder s iequality, which geeralizes the Cauchy-Schwarz iequality (this is visible from the = case); the first solutio above is adapted from the stadard proof of Hölder s iequality. We do t kow whether the declaratio Apply Hölder s iequality by itself is cosidered a acceptable solutio to this problem. 4.[3-A4] We split ito three cases. Note first that A a, by applyig the coditio for large x. Case : B 4AC >. I this case Ax +Bx+C has two distict real roots r ad r. The coditio implies that ax + bx + c also vaishes at r ad r, so b 4ac >. Now B 4AC = A (r r ) a (r r ) = b 4ac. Case : B 4AC ad b 4ac. Assume without loss of geerality that A a >, ad that B = (by shiftig x). The Ax + Bx + C ax + bx + c for all x; i particular, C c. Thus 4AC B = 4AC 4ac 4ac b.

4 Alterate derivatio (due to Robi Chapma): the ellipse Ax + Bxy + Cy = is cotaied withi the ellipse ax + bxy + cy =, ad their respective eclosed areas are π/(4ac B ) ad π/(4ac b ). Case 3: B 4AC ad b 4ac >. Sice Ax + Bx + C has a graph ot crossig the x-axis, so do (Ax + Bx + C) ± (ax + bx + c). Thus ad addig these together yields Hece b 4ac 4AC B, as desired. (B b) 4(A a)(c c), (B + b) 4(A + a)(c + c) (B 4AC) + (b 4ac). 5.[3-B6] First solutio: (composite of solutios by Feg Xie ad David Pritchard) Let µ deote Lebesgue measure o [, ]. Defie E + = {x [, ] : f(x) } E = {x [, ] : f(x) < }; the E +, E are measurable ad µ(e + ) + µ(e ) =. Write µ + ad µ for µ(e + ) ad µ(e ). Also defie I + = f(x) dx E + I = f(x) dx, E so that f(x) dx = I + + I. From the triagle iequality a + b ±( a b ), we have the iequality f(x) + f(y) dxdy ± ( f(x) f(y) )dxdy = ±(µ I + µ + I ), E + E E + E ad likewise with + ad switched. Addig these iequalities together ad allowig all possible choices of the sigs, we get f(x) + f(y) dxdy max {, (µ I + µ + I ), (µ + I µ I + )}. (E + E ) (E E +) To this iequality, we add the equalities to obtai f(x) + f(y) dxdy E + E + f(x) + f(y) dxdy = µ + I + E E f(x) + f(y) dxdy = µ I f(x) dx f(x) dx = (µ + + µ )(I + + I ) max{(µ + µ )(I + + I ) + µ (I I + ), (µ + µ )(I + I ), (µ µ + )(I + + I ) + µ + (I + I )}. Now simply ote that for each of the possible comparisos betwee µ + ad µ, ad betwee I + ad I, oe of the three terms above is maifestly oegative. This yields the desired result.

5 Secod solutio: We will show at the ed that it is eough to prove a discrete aalogue: if x,...,x are real umbers, the x i + x j x i. i,j= I the meatime, we cocetrate o this assertio. Let f(x,...,x ) deote the differece betwee the two sides. We iduct o the umber of ozero values of x i. We leave for later the base case, where there is at most oe such value. Suppose istead for ow that there are two or more. Let s be the smallest, ad suppose without loss of geerality that x = = x a = s, x a+ = = x a+b = s, ad for i > a + b, either x i = or x i > s. (Oe of a, b might be zero.) Now cosider a terms b terms {}}{{}}{ f( t,, t, t,, t, x a+b+,, x ) as a fuctio of t. It is piecewise liear ear s; i fact, it is liear betwee ad the smallest ozero value amog x a+b+,..., x (which exists by hypothesis). Thus its miimum is achieved by oe (or both) of those two edpoits. I other words, we ca reduce the umber of distict ozero absolute values amog the x i without icreasig f. This yields the iductio, pedig verificatio of the base case. As for the base case, suppose that x = = x a = s >, x a+ = = x a+b = s, ad x a+b+ = = x =. (Here oe or eve both of a, b could be zero, though the latter case is trivial.) The f(x,...,x ) = s (a + b + (a + b)( a b)) s (a + b) = s (a ab + b ). This proves the base case of the iductio, completig the solutio of the discrete aalogue. To deduce the origial statemet from the discrete aalogue, approximate both itegrals by equallyspaced Riema sums ad take limits. This works because give a cotiuous fuctio o a product of closed itervals, ay sequece of Riema sums with mesh size tedig to zero coverges to the itegral. (The domai is compact, so the fuctio is uiformly cotiuous. Hece for ay ǫ > there is a cutoff below which ay mesh size forces the discrepacy betwee the Riema sum ad the itegral to be less tha ǫ.) Alterate derivatio (based o a solutio by Da Berstei): from the discrete aalogue, we have i<j f(x i ) + f(x j ) f(x i ), for all x,..., x [, ]. Itegratig both sides as (x,..., x ) rus over [, ] yields ( ) f(x) + f(y) dy dx ( ) f(x) dx, or f(x) + f(y) dy dx f(x) dx. Takig the limit as ow yields the desired result. Third solutio: (by David Savitt) We give a argumet which yields the followig improved result. Let µ p ad µ be the measure of the sets {x : f(x) > } ad {x : f(x) < } respectively, ad let µ / be mi(µ p, µ ). The f(x) + f(y) dxdy ( + ( µ) ) f(x) dx.

6 Note that the costat ca be see to be best possible by cosiderig a sequece of fuctios tedig towards the step fuctio which is o [, µ] ad o (µ, ]. Suppose without loss of geerality that µ = µ p. As i the secod solutio, it suffices to prove a stregtheed discrete aalogue, amely ( a i + a j + i,j ( p ) ) ( ) a i, where p / is the umber of a,..., a which are positive. (We eed oly make sure to choose meshes so that p/ µ as.) A equivalet iequality is i<j a i + a j ) ( p + p a i. Write r i = a i, ad assume without loss of geerality that r i r i+ for each i. The for i < j, a i + a j = r i + r j if a i ad a j have the same sig, ad is r i r j if they have opposite sigs. The left-had side is therefore equal to ( i)r i + r j C j, where to C j = #{i < j : sg(a i ) = sg(a j )} #{i < j : sg(a i ) sg(a j )}. Cosider the partial sum P k = k j= C j. If exactly p k of a,..., a k are positive, the this sum is equal ( ) ( ) [( ) ( ) ( )] pk k pk k pk k pk +, which expads ad simplifies to j= p k (k p k ) + ( ) k. For k p eve, this partial sum would be miimized with p k = k, ad would the equal k ; for k < p odd, this partial sum would be miimized with p k = k± k, ad would the equal. Either way, P k k. O the other had, if k > p, the p k (k p k ) + ( ) k p(k p) + sice p k is at most p. Defie Q k to be k if k p ad p(k p) + ( k ) if k p, so that Pk Q k. Note that Q =. Partial summatio gives r j C j = r P + j= r Q + = (r j r j )P j j= (r j r j )Q j j= r j (Q j Q j ) j= = r r 4 r p j=p+ ( ) k (j p)r j.

7 It follows that i<j a i + a j = ( i)r i + p r j C j j= ( i [i eve])r i + = ( p) ( p) ( p) r i + p p r i + p r i + p p i=p+ ( p)r i (p + i [i eve])r i r i r i, as desired. The ext-to-last ad last iequalities each follow from the mootoicity of the r i s, the former by pairig the i th term with the (p + i) th. Note: Compare the closely related Problem 6 from the USA Mathematical Olympiad: prove that for ay oegative real umbers a,..., a, b,..., b, oe has mi{a i a j, b i b j } i,j= mi{a i b j, a j b i }. i,j= 7.[996-B] By estimatig the area uder the graph of lx usig upper ad lower rectagles of width, we get lxdx (l(3) + + l( )) 3 + lxdx. Sice lxdx = xlx x + C, we have, upo expoetiatig ad takig square roots, ( e ) < ( ) e + 3 ( ) ( + ) + ( + < e ) + e + 3 3/, usig the fact that < e < 3. 8.[996-B3] View x,..., x as a arragemet of the umbers,,..., o a circle. We prove that the optimal arragemet is..., 4,,,, 3,... To show this, ote that if a, b is a pair of adjacet umbers ad c, d is aother pair (read i the same order aroud the circle) with a < d ad b > c, the the segmet from b to c ca be reversed, icreasig the sum by ac + bd ab cd = (d a)(b c) >. Now relabel the umbers so they appear i order as follows:...,a 4, a, a =, a, a 3,...

8 where without loss of geerality we assume a > a. By cosiderig the pairs a, a ad a, a 3 ad usig the trivial fact a > a, we deduce a > a 3. We the compare the pairs a 4, a ad a, a 3, ad usig that a > a, we deduce a 3 > a 4. Cotiuig i this fashio, we prove that a > a > > a ad so a k = k for k =,,...,, i.e. that the optimal arragemet is as claimed. I particular, the maximum value of the sum is + ( ) ( ) = + + ( ) + + [( ) ] = ( )( ) + ( ) + 6 = Alterate solutio: We prove by iductio that the value give above is a upper boud; it is clearly a lower boud because of the arragemet give above. Assume this is the case for. The optimal arragemet for is obtaied from some arragemet for by isertig betwee some pair x, y of adjacet terms. This operatio icreases the sum by x + y xy = ( x)( y), which is a icreasig fuctio of both x ad y. I particular, this differece is maximal whe x ad y equal ad. Fortuately, this yields precisely the differece betwee the claimed upper boud for ad the assumed upper boud for, completig the iductio. 9.[99-A5] If u, v, the u + v u + uv + v = u+v. Takig u = x ad v = y y we obtai y y x4 + (y y ) dx (x + (y y ))dx = y 3 y3 3 with equality everywhere if y = : the last iequality uses the positivity of d dy (y 3 y3 ) = y( y) o (, ). Thus the maximum value is 3..[988-B] First solutio: If y, the y(y ) x as desired, so assume y >. If x y / the If x y / the y( y) = y(y + ) y (x + ) y = x + x + y x. x y y + > y(y ). 4 Secod solutio: As i First solutio, we may assume y >. We are give y(y + ) (x + ), so x + y(y + ) ad x y(y + ) y(y ). (The last iequality follows from takig a = y ad b = y i the iequality (a + )(b + ) ab + for a, b >, which is equivalet (via squarig) to a + b ab, the AM-GM Iequality.) Squarig gives the result..[978-a5] Let g(x) = l ( ) si x x = l(six) lx. The g (x) = si x + x < for < x < π sice x > six for x >. Thus the graph of g(x) is cocave dow ad hece ( g(x i ) g x ) i = g(x), or g(x i ) g(x). Sice e x is a icreasig fuctio, this implies six ( ) i si x = e g(xi) e g(x) =. x i x