SOME NOTES ON INEQUALITIES

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Malcolm Kory Owens
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1 SOME NOTES ON INEQUALITIES Rihard Hoshio Here are four theorems that might really be useful whe you re workig o a Olympiad problem that ivolves iequalities There are a buh of obsure oes Chebyheff, Holder, Mikowski, Youg, et), whih are virtually ever appliable, so we ll just stik to four You ll see that whe you orretly use oe of these theorems, a brutally tough questio a reveal itself to have a elegat si-lie solutio The Fab Four of Iequality Theorems: Well, really the Fab Three, sie ) is just a speial ase of ), but oh well It s all good AM-GM arithmeti mea geometri mea iequality) If a,a,,a are o-egative real umbers, the a a,a with equality ourrig if ad oly if }{{} iff a,a a, arithmeti geometri mea mea a = a = = a Cauhy or Cauhy-Shwarz-Bujakovsky if you really wat to get tehial) If a,a,,a,b,b,,b are o-egative real umbers, the a a a )b b b ) a b a b a b ) with equality ourrig iff a = a = = a b b b Power-Mea a Let f) = a ) / a, where a,a,,a are o-egative real umbers, ad Suppose ad y are itegers with y The, f) fy), with equality ourrig iff a = a = = a For eample, the QM-AM-GM-HM iequality is derived from the Power Mea, sie f) f) f0) f ) a a a a a a a a a quadrati mea Atually, f0) does t eist, but as 0, f) approahes a a a a a a harmoi mea 4 Jese s Iequality This is the really ool oe Suppose that f) is a real otiuous futio that is ove also alled oave up) o a iterval You test for oveity by showig that f ) 0

2 for all i that iterval the double or seod derivative if you ve ever take alulus A easier way to see it is if you pik ay two poits o the urve, ad joi them If the lie lies above or o) the urve, it is ove If the lie lies below, the it is oave, also alled oave dow see diagram) Let a,a,,a be real umbers i a iterval S where f) is ove for all i S The, fa )fa ) fa ) with equality ourrig if ad oly if a = a = = a If f) is oave for all i S, the fa )fa ) fa ) ) a a a f a a a f Same thig, eept ow the sig is reversed Same oditios for equality as well ie, a = a = = a ) These might be quite ofusig, espeially the last oe, so let s try some eamples of how we a use these iequalities to solve really hallegig problems The last four are from reet Olympiad otests Just for Fu, Jeses for =: a a C, fa ) )fa ) ) Ba,fa )) Aa,fa )) a a D, fa ) a )) Let Aa,fa )), ad Ba,fa )) be ay two poits o the graph of the futio f), where the futio is ove Let C be the midpoit of the lie AB, ad D be the poit idiated The -oordiates of C ad D are the same, but C is above D, ie equality ours iff A = a, ie A ad B are the same poit fa )fa ) a a f )

3 What is the miimum possible positive value of 9? Clearly, we do t wat to be egative If >0, the both ad 9 AM-GM iequality, 9 9 = 9= are positive, so by the Thus, 9 6, with equality ourrig iff = 9, ie, 9or = Note:, sie is positive) We see that, ideed, whe =, we have 9 ==6,thusthe miimum possible value of 9 is 6 Prove that a b ) a b ) 9 if a, b, > 0 Solutio : Epadig, we have a b ) a b ) = a b a b a b a b By AM-GM, sie a, b, > 0, we have a b a b a b a b 6 6 a b a b a b a b = Thus, a b a b a b a b 6, ad so a b ) a b ) 6 = 9, as required Solutio : By Cauhy-Shwarz, let a = a, a = b, a =, b =,b =,b =, a b ad thus: a b ) a b ) ) = =9, ad we are doe Equality ours iff a = a = a, ie iff a = b = b b b Fid the maimum value of 4 ), where 0 <<4 Yuk How a we use our kowledge of iequalities here? No really obvious way That s why a little trikery is eeded: Really ie trik here See how the left side simplifies iely? Sie 0 <<4, we have by AM-GM, 4 ) ) 4 4 ) 4 4 ) 4 = 4 4 ) 7 4 ) 7 Ma value is 7 Equality ors iff = = =4, ie if 4 =4, or = Chekig, we see that if =, the maimum value of 7 is ideed attaied

4 4 If a b =, show that a a) b ) ) 00, where a, b, > 0 b By Cauhy, [ a ) b ) ) ] [ ] a b = = [ a ) b ) )] a b a b a b ) a b ), sie a b = Usig Cauhy agai, we have Cauhy here?) Sie a b =, that meas a b 9 [ Hee, a a a a) b ) b a b ) a b ) ) = 9 See how we used ) b ) ) ] b a b ) 9) = 00, thus ) 00, as required 5 If the roots of the polyomial a 4 b d are all positive, fid a, b,, ad d There are two thigs you should be thikig: i) how is this a iequality problem? ad ii) surely there is t eough iformatio to figure this out! Chek this out: Let the { roots of the polyomial be p,p,p,p 4,p 5 ad p 6 We are give that p,p,,p 6 > 0 p p Also, p p 4 p 5 p 6 =6 relatioship betwee the roots of a polyomial ad p p p p 4 p 5 p 6 = its oeffiiets) By the AM-GM iequality, p p p p 4 p 5 p p p p p 4 p 5 p 6 works oly beause all the terms are o-egative) Whoa, we have equality, ie = That tells us that p = p = p = p 4 = p 5 = p 6! from AM-GM Sie p p p 6 = 6, that tells us that eah term is equal to Hee all the roots of the polyomial are, so: a 4 6 d = ) 6 = ad mathig up oeffiiets, we get a =5,b=0,= 5 ad d = 6 Is t that a ool questio? 4

5 6 Let A, B, ad C be the agles of a triagle Show that si A si B si C Here s where our buddy Jese omes i hady Let f) = si The f ) = os, ad f ) = si Sie A, B, ad C are agles of a triagle, 0 < A,B,C < 80 Also, A B C = 80, but we ll use that later For all from 0 to 80, f ) < 0, ad thus is oave i that iterval Hee, sie A, B, ad C lie i this iterval, by Jese s Iequality fa)fb)fc) si A si B si C ) A B C f ) 80 si =, sie A B C = 80 Multiplyig both sides of the iequality by, we arrive at the desired result 7 Let a, b, ad be positive real umbers Show that a a b b ab) ab/ 995 CMO) There are several ways to do this, but this oe is really igeious) Let f) = l ) The f ) =l ) =l, ad f ) = ad f ) > 0 for all positive values of This, by Jese s Iequality, for positive a, b, ad, we have: ) fa)fb)f) a b a l a b l b l a b ) a b l ) a b ab l aa lb b l ) l ) a b ab l a a b b l ) a b ab a a b b a b ab,thusa a b b Thus, we have prove the desired iequality Note: equality ours iff a = b = By AM-GM, a b ) ab ab) ab =ab) ab 8 Suppose a, b, ad are the sides of a triagle Show that: a b b a a b a b 996 APMO, last questio) a b > 0 Sie a, b, ad are the sides of a triagle a b > 0 Let = ab, y = a = b b a > 0 ad z = b a The, y, z > 0 ad we a epress a, b, ad i terms of, y, ad z Our iequality the beomes, ie is equivalet to): y z y 5 z y z

6 Sie, y, z > 0, ad thus,, y, z>0 So let s make aother substitutio: = p,y = q, ad z = r, where p, q, r > 0 The we eed to prove that: p q r p q q r q r p But by GM-AM, we have q p q p, r p r, ad Really eat use of symmetry here Always try to eploit symmetry) Addig up these three iequalities, we get q r q r p q p r q r p q p r q r = p q r as desired Thus, we are doe 9 Suppose a, b, ad are all positive Prove that a b ab) b ab) a ab) ab) 998 USAMO, Questio # ) Here s a really ie trik to remember: if we replae a, b, ad by ka, kb, ad k, all the terms with k s will ael out, ad we ll get bak to the origial iequality try it, you ll see that all the k s will disappear Thus, we a assume without loss of geerality that ab =!!! It makes thigs so muh easier! Eve if a, b ad are t umbers that multiply to, we a multiply all of them by a ostat k so that the relatio holds, so that s why we a do that Furthermore, we a let = a,y= b, ad z =, sie that will make the simplifiatio easier Sie ab =, we have yz = a b = So ow our iequality beomes: y ) y z ) z ), where, y, z > 0 ad yz = Muh easier, is t it? Well, after simplifiatio, we get: y z) y y z z yz y z = y z)y yz z) yz y z)y yz z ) yz = Ad beause y z y yz z yz = ad y z =, by AM-GM, we have y z)y yz z ) ) =, ad so we are doe 0 Suppose that a, b, ad are positive real umbers suh that ab = Prove that: a b ) b a ) a b) 995 IMO, Questio # ) The trik is to make the substitutio a =,b= y, ad = If you do t do this, the z problem is virtually impossible to solve See how that trik is useful? Try to reogize substitutios like these that will give you a iequality that is easier to prove Remember, whe you see a iequality problem, be lever ad try to use some of the ideas detailed i these solutios Who kows, that might be the way to do it! With problems like these, perserverae ad teaity is what you eed you might have to try 5 to 0 ore more!) differet methods before you fially get it! Thus, we have y z ) y z ) z y ) 6

7 yz y z y z z z y y y z y z z y sie yz = beause yz = ab =) Hee, if we a prove this iequality, we will be doe There are ow a ouple of ways to proeed Method : Cauhy) Sie, y, z 0, we have by Cauhy: y z y z y y z y z y z y z ) y z)z z) y)) y z) See how Cauhy is used here? Beautiful trik eploits the symmetry of the epressio) ) z y z) y z) y z y y z, by AM-GM y z yz =, so y z by AM GM Thus, we are doe Method : Jese) p Let fp) = y z p The oe a show that f p) = y z p p y z p) y z p), ad this is learly positive if 0 <p< y z Thus, sie 0 <,y,z < y z, by Jese, we have: f)fy)fz) ) y z y z y z z y yz ) y z) yz = y z By AM-GM, y z same as before), so y z y z z y, as required 7

8 SOME FUN PROBLEMS FOR YOU TO DO If a,a,a are o-egative, show that a a a 6 a/ a / a /6 Show that if a, b, > 0, the a b)a )b ) 8ab Fid the greatest value of y z give that, y, z > 0 ad y z =6 4 Show that if a, b, > 0, the a b ) a b a ) b our? 9 Whe does equality 5 If a, b > 0, prove that a b ab, where is a positive iteger Usig this, a you show that the sequee ) is ireasig? ) 6 Show that for all >, we have >! Note: We ever have equality) 7 a) Prove that amog all retagles with a fied perimeter p, the square has the greatest area b) Prove that amog all retagles with a fied area A, the square has the least perimeter 8 Prove that amog all triagles with a fied perimeter p, the equilateral triagle has the greatest area 9 Solve the system: where, y, z are positive real umbers { yz = 64 yz = 8, 0 If A, B ad C are the agles of a triagle, show that os A os B os C Suppose that a,a,,a > 0 Show that a a a a a a a a a a a a Suppose a,a,,a 999 are 999 positive real umbers, ad let b,b,,b 999 be some rearragemet of these umbers What is the miimum value of a a a 999? b b b 999 8

9 Suppose that a a a > 0, ad that b a,b b a a,b b b a a a,,b b b a a a Prove that b b b a a a quite hard!) 4 Let be a iteger, Let a,a,,a be real umbers, where a i for i =,,,Ifs = a a a, prove that: 995 IMO Shortlist Problem) a a a a a a a a a 4 a a a 4 a a a a a a s 5 Prove the AM-GM iequality usig Jese s Iequality Hit: Let f) = l ) If you have ay questios, please feel free to me! rhashio@udergradmathuwaterlooa 9

Class #25 Wednesday, April 19, 2018

Class #25 Wednesday, April 19, 2018 Cla # Wedesday, April 9, 8 PDE: More Heat Equatio with Derivative Boudary Coditios Let s do aother heat equatio problem similar to the previous oe. For this oe, I ll use a square plate (N = ), but I m