Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207
Contents Wrm up nd Am-Gm inequlity 2. Elementry inequlities...................... 2.2 Aritmeti Men Geometri Men.............. 2.3 Exmples............................. 2 2 Common identities nd other mens 4 2. Identities.............................. 4 2.2 Other Mens: Qudrti Men nd Hrmoni Men...... 4 2.3 Exmples............................. 5 3 Geometri (tringle) inequlities 6 3. Nottion.............................. 6 3.2 The si tringle inequlity nd ommon sustitution... 6 4 Cuhy-Shwrtz, Titu s lemm nd Nesitt s inequlity 8 4. Cuhy-Shwrtz inequlity................... 8 4.2 Titu s lemm........................... 8 4.3 Nesitt s inequlity........................ 9 4.4 Exmples............................. 0 5 Cheyshev inequlity 5. Oriented sequenes nd Cheyshev inequlity......... 5.2 Exmples............................. 2
Astrt This ooklet is designed to riefly summrize mth ourse offered during the Spring HSSP 207 orgnized y Edutionl Studies Progrm t Msshusetts Institute of Tehnology. It is designed for students with no prior experiene with Olympid Inequlities who hve fundmentl knowledge of middle shool lger. The min gol is to introdue the students with this ommon Mth Olympid topi nd present them with retive nd elegnt methods of solving prolems of similr kind.
Wrm up nd Am-Gm inequlity. Elementry inequlities if nd then for ny,, R if then + + for ny,, R if nd x y then + x + y for ny,, x, y R if nd x y then x y for ny,, x, y R +.2 Aritmeti Men Geometri Men Let, 2..., n e non-negtive rel numers. Then, the following inequlity holds: + 2 +... + n n n 2 n () Equlity holds for = 2 =... = n.3 Exmples Exmple Prove the following inequlity for every,, R + : Applying AM-GM inequlity on 2 2 2 2 + 2 2 + 2 2 + + nd 2 2 gives us: 2 + 2 2 2 2 2 2 2 = 2 2 This is wesome, euse is one of the terms on the RHS of the inequlity. Completely nlogously we oserve: 2 + 2 2 2 2 2 2 2 = 2 2 2
nd 2 + 2 2 2 2 2 2 2 = 2 2 If we sum up these inequlities nd divide y 2 we hve: 2 2 + 2 2 + 2 2 + + QED Note tht equlity ours when = = Exmple 2 Prove the following inequlity for every,, R + : When does equlity our? Using AM-GM inequlity we hve: ( + )( + )( + ) 8 + 2, + 2 nd + 2 Therefore, ( + )( + )( + ) 8 = 8 nd tht is wht we needed to prove. Equlity ours when =, = nd = leding to = = = Exmple 3 Find ll rel solutions of the following system: x + y = 2 xy z 2 = Even though it might seem tht this prolem hs nothing to do with inequlities, we will show the power of using inequlities to solve mny kinds of prolems. Let s strt y noting tht 2 = x + y 2 xy. By doing some simple lger, it is not hrd to onlude tht xy. On the other hnd, using 3
the seond eqution nd the ft tht xy we get: = xy z 2 z 2. By nelling out the ones on oth sides, we get tht 0 z 2 Sine every numer squred is greter or equl to 0, it follows from the lst inequlity tht z must e 0. Therefore, we lso get tht equlity holds for x = y =. This is one exmple where inequlities n e used to solve other types of prolems. 2 Common identities nd other mens 2. Identities There re mny identities tht prolem solvers use in order to prove inequlities. They llow us to trnsform the inequlity to nother, equivlent inequlity whih is esier to prove. Here re some of the most used identities: 2 + 2 + 2 = ( + + ) 2 2( + + ) ( + )( + )( + ) = ( + + )( + + ) ( + )( + )( + ) = + ( + + ) + ( + + ) + (+)(+)+(+)(+)+(+)(+) = (++) 2 +(++) ( + ) + ( + ) + ( + ) = ( + + )( + + ) 3 2.2 Other Mens: Qudrti Men nd Hrmoni Men Until now we hve fmilirized ourselves with Arithmeti nd Geometri Mens nd the inequlity tht holds etween them. Now we will introdue nother two mens nd we will see how they fit into the so-lled Mens inequlity. The Qudrti Men for n numers is defined s: Q = 2 + 2 2 +... + 2 n n 4
The Hrmoni Men for n numers is defined s: n H = + 2 +... + n If we denote Arithmeti Men nd Geometri Men with A nd G respetively, we n introdue the following importnt result: Mens Inequlity: Q A G H Equlity holds for = 2 =... = n 2.3 Exmples Exmple 4 Let,,, d > 0 suh tht + + + d =. Prove the inequlity: 4 + 3 + + 3 + + 4d + + 4 + 3d + 4 + 3 + d 2. Let A = nd D = 4 + 3 + d.. Using Am-Hm for A,B,C,D we get: = A + B + C + D 4 + 3 +, B = 3 + + 4d, C = + 4 + 3d 6 A + B + C + D 4 4 + + + A B C D + + + A B C D 6 LHS (4 + 3 + ) + (3 + + 4d) + ( + 4 + 3d) + (4 + 3 + d) 6 8 + 8 + 8 + 8d = 6 8 = 2 Exmple 5 Let,,, > 0 suh tht =. Prove the inequlity: ( + )( + ) + ( + )( + ) + ( + )( + ) 3 4 = 5
One of the useful strtegies here would e to get rid of the denomintors nd hope to get something etter. Let s multiply everything with 4(+)(+)(+ ). We get the following: 4( + ) + 4( + ) + 4( + ) 3( + )( + )( + ) whih is equivlent to: 4( + + + + + ) 3( + + + + + + + ) Note tht here we used one of the ove mentioned identities to simplify the expression. By nelling the like-terms we otin tht we re left to prove: + + + + + 3( + ). If we sustitute = (whih is given in the sttement of the prolem) we re left to prove tht + + + + + 3( + ) = 6. Note tht the lst inequlity is true from AM-GM inequlity pplied on the six terms,,,,, nd using one more the ft tht =. 3 Geometri (tringle) inequlities 3. Nottion There re some ommon nottions when it omes to tringles:,, - sides of the tringle A - Are s- hlf-perimeter, nmely s = ++ 2 R - irumrdius (rdius of the irumirle) r - inrdius (rdius of the inirle) 3.2 The si tringle inequlity nd ommon sustitution One of the most importnt things to note out tringles is tht the sum of ny two sides is ALWAYS greter thn the third side. Tht mens, if,, 6
re sides of tringle, then ll of the following inequlities hold: + > + > + > Another thing tht we will often enounter while solving tringle inequlities is the ft tht the sides of ny tringle n e represented in the following wy: = x + y, = x + z, = y + z where x, y, z re positive rel numers. This is lled Rvi trnsformtion nd it holds true sed on the properties of the insried irle in tringle, s illustrted in figure. Figure : Tringle nd its inirle. Exmple 6 Let,, e sides of tringle. Prove the following inequlity: + + + + + < 2 Sine,, re sides of tringle we hve tht + >. If we dd + on oth sides of the inequlity we get 2( + ) > + +. This is equivlent to + > ++ = s i.e. <. By multiplying this with we get <. 2 + s + s This looks extly like the third term on the left hnd side of the inequlity. Note tht nlogously we n get the other two terms. If we dd these results we get tht + + + + + < + + s whih is extly wht we wnted to prove. 7 = + + ++ 2 = 2
Exmple 7 Let,, e sides of tringle. Prove the following inequlity: ( + )( + )( + ) Using the Rvi trnsformtion we n mke the following sustitution: = x + y, = y + z, = z + x. Note tht even though this is not identil s the sustitution written ove, the essene nd mening re ompletely the sme. Note tht from this we get: + = 2y, + = 2z nd + = 2x. Now the initil inequlity looks like this: (x + y)(y + z)(z + x) 8xyz This proof of this lst inequlity is strightforwrd pplition of AM-GM inequlity in eh of the prentheses nd multiplying them together, similr s in exmple 2. 4 Cuhy-Shwrtz, Titu s lemm nd Nesitt s inequlity 4. Cuhy-Shwrtz inequlity Cuhy-Shwrtz is one of the most ommon inequlities esides AM-GM. It is stted s follows: Let, 2..., n nd, 2..., n e rel numers. Then the following holds: ( 2 + 2 2 +... + 2 n)( 2 + 2 2 +... + 2 n) ( + 2 2 +... + n n ) 2 with equlity if nd only if = 2 2 =... = n n While we will not give full proof of this theorem in this osion, we highly enourge the reder to tryto prove the sttement for n = 2. 4.2 Titu s lemm Titu s lemm is diret onsequene of Cuhy-Shwrtz inequlity. It is lso known s Cuhy-Shwrtz inequlity in Engel s form. It is n extremely 8
useful shortut for solving inequlities whose numertors re perfet squres. Let s tke look t it: Let, 2..., n nd, 2..., n e rel numers. Then the following holds: 2 + 2 2 +... + 2 n ( + 2 +... + n ) 2 2 n + 2 +... + n Equlity holds for = 2 2 =... = n n The proof of this inequlity is strightforwrd onsequene of Cuhy- Shwrtz inequlity, whih is esily oserved if we multiply oth sides of the inequlity with + 2 +... + n. 4.3 Nesitt s inequlity Nesitt s inequlity deserves speil ple in the world of olympid inequlities sine it often ppers s prt of proving lrger prolems. Let s tke look t it: Let,, e positive rel numers. Then we hve: + + Equlity is hieved for = = + + + 3 2 Proof: There re mny wys to prove this inequlity, ut here we will mke use of Cuhy-Shwrtz inequlity. It would e nie if we n mke ll numertors the sme. To hieve this we will dd the numer to eh frtion seprtely, whih is equivlent to dding 3 to oth sides of the inequlity. This yields to: + + + Note tht this is equivlent to: + + + + + 9 2 + + + + + + + + + + + 9 2 9
or 2( + + )( + + + + + ) 9 whih n lso e written s (( + ) + ( + ) + ( + ))( + + + + + ) 9 Note tht if we tret + s, + s in the definition ove nd nlogously with the other ones, we n pply Cuhy-Shwrtz inequlity esily to get tht the left hnd side of the lst inequlity is greter thn or equl to ( + + ) 2 whih is just 9 s we needed to prove. Note: We ould hve proven the lst inequlity very esily using AM-HM! Think out how would you do tht. 4.4 Exmples Exmple 8 Let,, e rel numers. Prove the inequlity: 2 2 + 3 2 + 6 2 ( + + ) 2 While the numers on the left hnd side seem unrelted nd intimidting, they hve one property whih immeditely solves the inequlity with the help of Cuhy-Shwrtz inequlity. Nmely + + =. Now we hve the 2 3 6 following: LHS = (2 2 + 3 2 + 6 2 ) = ( 2 + 3 + 6 )(22 + 3 2 + 6 2 ) ( + + ) 2 nd tht is wht we needed to prove. Exmple 9 Let x, y, z e positive rel numers. Prove the following inequlity: 2 x + y + 2 y + z + 2 z + x 9 x + y + z If we look losely t the denomintors, we might e le to see tht pplying Titu s lemm should e helpful in this se. The ft tht the nomintors 0
re not perfet squres shouldn t disourge us, sine we n mke them to e perfet squres! We hve the following: ( 2) 2 x + y + ( 2) 2 y + z + ( 2) 2 z + x (3( 2)) 2 2(x + y + z) = 9 x + y + z whih is the right hnd side of the inequlity. 5 Cheyshev inequlity 5. Oriented sequenes nd Cheyshev inequlity This is nother very useful nd ommonly used inequlity whih n e pplied on oriented sequenes (monotonilly inresing or deresing). There re two forms of this inequlity, one for similrly oriented sequenes, nd one for oppositely oriented sequenes. By similrly oriented we men tht the sequenes re either oth inresing or oth deresing. We will look t the se when they re oth deresing. Let 2... n nd 2... n e rel numers. Then we hve: n( + 2 2 +... + n n ) ( + 2 +... + n )( + 2 +... + n ) Equlity is hieved when either of the sequenes is onstnt. Let s see the se when the sequenes re oppositely oriented. Let 2... n nd 2... n e rel numers. Then we hve: n( + 2 2 +... + n n ) ( + 2 +... + n )( + 2 +... + n ) Equlity is hieved when either of the sequenes is onstnt. As we n see, the only differene in the two theorems is the sign on the inequlity nd the orienttion of the sequenes. In mny of the inequlities we n ssume without loss of generlity (WLOG) tht the numers re ordered. In other words, ssuming tht would give us the sme
result s if we ssumed tht or ny other ordering. Let s see some exmples. 5.2 Exmples Exmple 0 Prove Nesitt s inequlity using Cheyshev inequlity. In se we forgot, let s remind ourselves of Nesitt s inequlity: + + + + + 3 2 Let s denote the sum + + to e S. Then we n write the inequlity s: S + S + S 3 2 Note tht we n ssume without loss of generlity tht. Then it is ler tht nd if we dd S to oth sides we get S S S. Note tht the sequenes nd S S S re oppositely oriented. Now here omes the power of Cheyshev inequlity for oppositely oriented sequenes: ( S + S + )((S ) + (S ) + (S )) 3( + + ) S Now, we hve tht (S ) + (S ) + (S ) = 2S = 2( + + ). The sum ( + + ) nels with the one in right hnd side nd we get the desired inequlity, nmely: S + S + or + + + + whih is wht we wnted to prove. S 3 2 + 3 2 Exmple Let,,, d e positive rel numers. Prove the following: 5 + 5 + 5 + d 5 d( + + + d) WLOG we n ssume d. From this immeditely follows tht 2
5 5 5 d 5. The importnt thing is tht these sequenes re similrly oriented (we ould hve ssumed tht oth of them re inresing insted). Then y Cheyshev inequlity nd AM-GM we get: 4( 5 + 5 + 5 +d 5 ) ( 4 + 4 + 4 +d 4 )(+++d) 4 4 4 4 4 d 4 (+++d) = 4d(+++d) If we nel the onstnt 4 on oth sides we get the desired inequlity, nmely: 5 + 5 + 5 + d 5 d( + + + d) 3
Conlusion We hope tht this ooklet ws helpful to etter understnd the onepts overed on letures nd to mke you see the euty nd the rt of prolem solving. However, we lso hope tht this is just eginning of your journey with olympid inequlities. There re mny online resoures tht n help you ontinue developing your knowledge of this topi. Just google it! Good luk! 4