100 INEQUALITY PROBLEMS ***** Cao Minh Quang 05/11/2006 Most of these problems were collected from the Mathematics and Youth Magazine in Vietnam. Translation from Vietnamese into English may have many errors. I am looking forward to hearing from your ideas. • Address: Cao Minh Quang, Mathematics teacher, Nguyen Binh Khiem specialized High School, Vinh Long town, Vinh Long, Vietnam. • Email: kt13quang@yahoo.com 1. () a,b 0,a b 1>+=, 22 ab4 a1b15 + ≤ ++ . First solution. Applying the AM – GM Inequality we get 22 2 2 1 3 1 3 4a 3 a 4a a1a 2a. 44 44 4 a 14a3 + += + + ≥ + = ⇒ ≤ + + . Similarly, 2 b 4b 4b 3 b1 ≤ + + . Adding these two inequalities, 22 ab 4a4b 1 1 23 a 1b 1 4a34b3 4a34b3 ⎡ ⎤ ⎛⎞⎛⎞ +≤ + =− + ⎜⎟⎜⎟ ⎢ ⎥ ++ ++ ++ ⎝⎠⎝⎠ ⎣ ⎦ . On the orther hand, ()() () 11 11 4 2 4a 3 4b 3 4 4a 3 4b 3 4a 3 4b 3 4 a b 6 5 ⎡⎤ ⎛⎞⎛⎞ ++ + + ≥⇒ + ≥ =⎡⎤ ⎜⎟⎜⎟ ⎢⎥ ⎣⎦ ++ ++ ++ ⎝⎠⎝⎠ ⎣⎦ . Thus, 22 ab 24 23. a1b1 55 + ≤− = ++ . Second solution. Applying the AM – GM Inequality we get 4 22 2 2 5 5 1111 1 5 a1a 5a 4a 4444 4 4 ⎛⎞ += ++++≥ = ⎜⎟ ⎝⎠ . Similarly, 22 5 5 b 14b 4 +≥ . Adding these two inequalities, 33 55 22 22 55 ab a b411 11 a. . b. . 55 a1b1 5 22 22 4a 4b 44 ⎛⎞ +≤ + = + ≤ ⎜⎟ ⎜⎟ ++ ⎝⎠ () 11 11 aaa bbb 3a b 2 444 22 22 55 5 555 ⎡⎤ ⎛⎞⎛⎞ ++++ ++++ ⎢⎥ ⎜⎟⎜⎟ ++ ⎡⎤ ≤+== ⎢⎥ ⎜⎟⎜⎟ ⎢⎥ ⎢⎥ ⎣⎦ ⎜⎟⎜⎟ ⎜⎟⎜⎟ ⎢⎥ ⎝⎠⎝⎠ ⎣⎦ . Third solution. Applying the AM – GM Inequality we get 1 ab2ab ab 4 + ≥⇒≤ . Therefore, () 22 2 22 2222 22 a b ab a b a b ab 1 a1b1aba b1 ab 2abab 1 +++ + += = = ++ +++ +−+ + 2 22 1 1 ab 1 ab 1 4 4 31 31 ab 2ab 2 5 1331 . ab ab 24 16 4216 + ++ == ≤= −+ ⎛⎞ −+ −−+ ⎜⎟ ⎝⎠ . 2. () a,b,c 0,a b c 1>++=, abc3 a1b1c1 4 + +≤ + ++ . First solution. Applying the AM – GM Inequality we get ()() aa1aa a1 a b ac 4ab ac ⎛⎞ =≤+ ⎜⎟ ++++ ++ ⎝⎠ , ()() b b1bb b 1ba bc4babc ⎛⎞ =≤+ ⎜⎟ ++++ ++ ⎝⎠ , ()() cc1cc c1 cb ca 4c b ca ⎛⎞ =≤+ ⎜⎟ ++++ ++ ⎝⎠ . Adding these three inequalities, abc1abbcac3 a1b1c1 4ab bc ac 4 +++ ⎛⎞ ++≤ ++ = ⎜⎟ +++ + ++ ⎝⎠ . Second solution. Applying the AM – GM Inequality we get +=+ + + ≥ ⇒ ≤ + 4 33 4 3 111 1 a 1 a1 a 4a a3 333 a14 3 , Similarly, 4 33 b1 b3 b1 4 ≤ + , 4 33 c1 c3 c1 4 ≤ + . Adding these three inequalities, 44 4 abc3 1 1 1 a.a.a. b.b.b. c.c.c. a1 b1 c1 4 3 3 3 ⎛⎞ ++≤ + + ≤ ⎜⎟ ⎜⎟ +++ ⎝⎠ 111 aaa bbb ccc 33 333 44 4 4 4 ⎡⎤ ⎛⎞⎛⎞⎛⎞ +++ +++ +++ ⎢⎥ ⎜⎟⎜⎟⎜⎟ ≤++= ⎢⎥ ⎜⎟⎜⎟⎜⎟ ⎢⎥ ⎜⎟⎜⎟⎜⎟ ⎢⎥ ⎝⎠⎝⎠⎝⎠ ⎣⎦ . Third solution. Applying the AM – GM Inequality we get ()()() ⎛⎞ ++ +++ + + ≥ ⎡⎤ ⎜⎟ ⎣⎦ +++ ⎝⎠ 111 a1 b1 c1 9 a1 b1 c1 ()()() ⎛⎞ ⇒++≥ = ⎜⎟ +++ +++++ ⎝⎠ 111 9 9 a1 b1 c1 a1 b1 c1 4 . Therefore, abc 111 93 33 a1 b1c1 a1 b1c1 4 4 ⎛⎞ ++=− ++ ≤−= ⎜⎟ +++ +++ ⎝⎠ . Forth solution Applying the AM – GM Inequality we get +=+ + ≥ + ⇒ ≤ + − + + 12 a2 a 3a11 1 a1 a 2 . 33 33 a1 2 23 a2 2 33 . Similarly, b3b111 . b1 2 2 3 b2 2 33 ≤+− + + , c3c111 . c1 2 2 3 c2 2 33 ≤+− + + . Adding these three inequalities, abc3abc311 1 1 a1 b1 c1 2 3 3 3 2 3 a2 b2 b2 222 33 33 33 ⎛⎞ ⎜⎟ ⎛⎞ ⎜⎟ ++≤ +++− + + ≤ ⎜⎟ ⎜⎟ +++ ⎜⎟ ⎝⎠ +++ ⎜⎟ ⎝⎠ 111 abc 33993 333 3 22 2 2 2 44 abc 2 23. 333 3 ⎡⎤ ⎛⎞⎛⎞⎛⎞ +++ ⎢⎥ ⎜⎟⎜⎟⎜⎟ ≤+++− ≤−= ⎢⎥ ⎜⎟⎜⎟⎜⎟ ⎛⎞ ⎢⎥ ⎜⎟⎜⎟⎜⎟ ++ + ⎜⎟ ⎢⎥ ⎝⎠⎝⎠⎝⎠ ⎣⎦ ⎝⎠ . 3. () a,b,c 1≥ , () 111 2abc 9 abc ⎛⎞ + ++ ≥ ⎜⎟ ⎝⎠ Solution. We have, () ( ) a1b1 0 ab1a b−−≥⇔+≥+, () ( ) ab 1 c 1 0 abc 1 ab c−−≥⇔+≥+. Adding these two inequalities, we obtain abc 2 a b c + ≥++. Thus, () () 111 111 2abc abc 9 abc abc ⎛⎞ ⎛⎞ +++≥++++≥ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ . 4. () x,y,z0,xyxyyzyzzxzx1>++=, + +≥ +++ 666 33 33 33 x y z1 x yy zzx2 . Solution. We set 333 a x , b y , c z=== and observe that ab bc ca 1 + +=. The inequality is equivalent to + +≥ +++ 222 abc1 ab bc ca 2 . Applying the Cauchy – Buniakowski Inequality we get ()()()( ) ⎡⎤ ++ +++++≥++ ⎡⎤ ⎢⎥ ⎣⎦ +++ ⎣⎦ 222 2 abc ab bc ca abc ab bc ca () () ⇒++≥++≥ ++= +++ 222 abc1 1 1 abc ab bc ca ab bc ca 2 2 2 . 5. () x, y 0,xy 1>= , ++++ ≥ ++ 22 22 9 x3xy3y 11 xy1 . Solution. Applying the AM – GM Inequality we get () () +++ ++−≥ ++ −= ++ ++ 22 22 4 22 22 99 x y 1 3x 3y 1 4 x y 1 . .3x.3y 1 11 xy1 xy1 . 6. { } () m, n \ 0 , m a, b, c n∈≤≤ , () () 2 nm abc3 bcacab 22mnm − ++≤+ +++ + . Solution. Without loss of generality, we can assume that abc≥≥ , we set xbc, yca, zab = +=+=+ . We observe that xyz≤≤, therefore, ⎛⎞⎛⎞ ⎛⎞ ⎛⎞ − −+− −≥ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠⎝⎠ yz xy 11 11 0 xy yz ⎛⎞ +++ ⇔++≤++ ⎜⎟ ⎝⎠ yz xz xy x z 22 xyz yx () () − ⎛⎞ ⇔+++≤ + ⎜⎟ +++ ⎝⎠ 2 zx abc 23261 bc ca ab xy . On the other hand, zx ac nm 0 xbc2m −−− ≤=≤ + and zx ac ac nc nm 0 zabacncnm − −−−− ≤=≤≤≤ + +++ . Thus, ()() () () −− ≤ + 22 zx nm 2 xz 2m n m . From (1) and (2), we have () () 2 nm abc3 bcacab 22mnm − ++≤+ +++ + . 7. () 12 n 0 x , x , , x 1, n 2<≤≥, ()()() 12 n 12 n xx x 1 n1n1x1x 1x +++ ≤ +− − − . Solution. We set () −= = ii 1 x a , i 1, 2, , n and observe that i 0a 1 ≤ < , ( ) =i 1, 2, , n . The inequality is equivalent to ()() ( ) 12 n 12 n 1a 1a 1a 1 n 1 na a a −+−++− ≤ + , or () ( ) 2 12 n 12n 12n a a a 1 a a a n a a a+++ + ≥ . Applying the AM – GM Inequality we get n 12 n 12n a a a n a a a+++≥ and () ( ) n1 n 12 n 12 n 12 n 1 na a a 1 n 1 a a a n a a a − +≥+− ≥ . Therefore, ()()()()() n 22 2 n 1 2 n 12n 12n 12 n 12n 12n a a a 1 a a a n a a a n a a a 1 a a a n a a a+++ + ≥ = +++ + ≥ . 8. () 0xyz1,3x2yz4<<≤≤ + +≤ , + +≤ 222 10 3x 2y z 3 . Solution. We have 3x 2y z 4++≤, thus, ++≤ 2 3x 2xy xz 4x . On the other hand, 0xyz1<<≤≤, therefore ( ) ( ) 2 yy x2 y x − ≤− and () zz x z x−≤− . Adding these two inequalities, we obtain + +≤++ 222 3x 2y z x 2y z . Applying the Cauchy – Buniakowski Inequality we get () () () ⎛⎞ ++ ≤++≤++ ++⇒++≤ ⎜⎟ ⎝⎠ 2 2 222 222 222 110 3x 2y z x 2y z 2 1 3x 2y z 3x 2y z 33 . 9. () 0xyz1≤≤≤≤, ()()() −+ −+ −≤ 222 108 xyz yzy z1z 529 . Solution. We set ()() ( ) =−+−+− 222 Txyz yzy z1z. Applying the AM – GM Inequality and the Cauchy – Buniakowski Inequality we get ()() () 3 22 11yy2z2y T0 y.y2z2y z1z z1z 223 ++ − ⎛⎞ ≤+ − + − ≤ + − ⎡⎤ ⎜⎟ ⎣⎦ ⎝⎠ 2 22 42354232323 zz1zz1z zz1z 27 27 23 54 54 27 ⎛ ⎞⎛ ⎞⎛⎞⎛⎞⎛⎞⎛ ⎞ =+−=−= − ⎜ ⎟⎜ ⎟⎜⎟⎜⎟⎜⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝⎠⎝⎠⎝⎠⎝ ⎠ 2 23 23 23 zz1z 54 108 54 54 27 23 3 529 ⎡⎤ ++− ⎢⎥ ⎛⎞ ≤= ⎢⎥ ⎜⎟ ⎝⎠ ⎢⎥ ⎣⎦ . 10. () 222 a,b,c ,abc8∈++≤ , + +≥−ab bc 2bc 8 . Solution. We have () 2 2 222 b3b 8 ab bc 2bc a b c ab bc 2bc a c 0 24 ⎛⎞ +++ ≥+++++ =+++ ≥ ⎜⎟ ⎝⎠ . Therefore, ab bc ca 8++≥−. 11. () a,b 0> , ()()() () () 2 22 ab 3aba3b ab ab 2 8a b a 6ab b −++ + ≥+ +++ . Solution. If ab= , the inequality is true. If ab≠ , the inequality is equivalent to ()()() () () −++ + −≥ +++ 2 22 ab 3aba3b ab ab 2 8a b a 6ab b () ( ) ()() () () ⎡⎤ +++ ⎢⎥ ⇔− − ≥ ⎢⎥ +++ ⎢⎥ ⎣⎦ 2 2 22 ab3aba3b ab1 0 4a b a 6ab b () ( ) ( ) ( ) ⇔+ ++−++ ++ ≥ 22 22 4 a b a 6ab b a b 2 ab 3a 3b 10ab 0 . We set xab, y ab=+ = and observe that 22 2 2 abx2y+=− . The inequality is equivalent to () () () () +−+ +≥⇔−≥⇔≥⇔+≥ 3 22 22 4x x 4 y x2 y 3x 4 y 0x2 y 0x2 y ab2ab. 12. () x ∈  , 33 sin x sin 2x sin3x 2 ++<. Solution. We have, ++=+ =+ ≤ 2 sin x sin 2x sin3x sin 2x 2sin 2x cos x sin 2x 4sin x cos x 22 2242 cos x cos x 1 4sinxcos x 1 4 sin xcos x 1 8 sin x. . 22 ≤+ ≤+ =+ 3 22 2 cos x cos x sin x 83 33 22 18 1 392 ⎛⎞ ++ ⎜⎟ ≤+ =+ < ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ . 13. () 12 n x , x , , x 0,n 3>≥, ()() ()() 22 22 123 234 n1 n 1 n 1 2 12 3 23 4 n1n 1 n1 2 xxx xxx xxx xxx n xx x xx x x x x x x x − − ++ ++ +++ +≥ ++ ++ . Solution. We set, ()() ()() 22 22 123 234 n1 n 1 n 1 2 12 3 23 4 n1n 1 n1 2 xxx xxx xxx xxx A xx x xx x x x x xx x − − ++ ++ =+++ + ++ ++ . Applying the AM – GM Inequality we have () () () () 22 22 123 234 n1 n 1 n 1 2 12 3 23 4 n1n 1 n1 2 xxx xxx xxx xxx A n 1 1 1 1 xx x xx x x x x x x x − − ⎡⎤⎡⎤⎡ ⎤⎡⎤ ++ ++ += ++ +++ ++ + ⎢⎥⎢⎥⎢ ⎥⎢⎥ ++ ++ ⎣⎦⎣⎦⎣ ⎦⎣⎦ ()() () ()() () ( ) ( ) () ()() () 1213 2324 n1nn11 n1n2 12 3 23 4 n1n 1 n1 2 xxxx xxxx x xx x xxxx xx x xx x x x x xx x −− − ++ ++ + + ++ =+ ++ + ++ + + ()( )( )( ) n 13 24 n11 n2 1324 n11n 2 n n 12 n1n 12 n1n 2xxxx xxxxx x x x x x x x nn2n x x x x x x x x − − −− ++ ++ ≥≥= . Therefore, ()() ()() 22 22 123 234 n1 n 1 n 1 2 12 3 23 4 n1n 1 n1 2 xxx xxx xxx xxx n xx x xx x x x x xx x − − ++ ++ +++ +≥ ++ ++ . 14. () ≤≤1a,b,c2, () 111 abc 10 abc ⎛⎞ + +++≤ ⎜⎟ ⎝⎠ . Solution. Without loss of generality, we can assume that 1abc2 ≤ ≤≤≤, we obtain, ab bc abbcac 11 11 0 2 bc ab bcabca ⎛⎞⎛⎞⎛⎞⎛⎞ − − +− − ≥⇔+++≤++ ⎜⎟⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ . Therefore, () 111 abbc ac ac abc 3 52 abc bcab ca ca ⎛⎞⎛ ⎞ ⎛⎞ ++ ++ =+ +++ ++≤+ + ⎜⎟⎜ ⎟ ⎜⎟ ⎝⎠⎝ ⎠ ⎝⎠ . On the other hand, ⎛⎞⎛ ⎞ ⎛⎞ ≤≤⇒ − − ≤⇒ +≤ ⇒+≤ ⎜⎟⎜ ⎟ ⎜⎟ ⎝⎠⎝ ⎠ ⎝⎠ 2 1a a1a a 5a ac5 12 0 1. 2c c2c c 2c ca2 . Thus, () 111 abc 10 abc ⎛⎞ + +++≤ ⎜⎟ ⎝⎠ . 15. () a,b,c,d 0> , 2222 22 22 4 abcdabcd bcda abcd + ++ +++≥ . Solution. Applying the AM – GM Inequality we get 222 6 8 222 222 4 abc a 8a 3. 2. 2 8 bcd bcd abcd +++≥ = , 222 6 8 222 222 4 bcd b 8b 3. 2. 2 8 cda cda abcd +++≥ = , 222 6 8 222 222 4 cda c 8c 3. 2. 2 8 dab dab abcd +++≥ = , 222 6 8 222 222 4 dab d 8d 3. 2. 2 8 abc abc abcd +++≥ = . Adding these four inequalities, we obtain 2222 22 22 444 a b c d abcd abcd abcd 68862 bcda abcd abcd abcd ⎛⎞ + ++ +++ +++ ⎛⎞⎛⎞⎛⎞ +++ +≥ = + ⎜⎟ ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠ ⎝⎠ 4 444 abcd 4abcd abcd 62.68 abcd abcd abcd +++ +++ ⎛⎞ ⎛⎞ ≥+=+ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ . Thus, 2222 22 22 4 abcdabcd bcda abcd + ++ +++≥ . 16. () 0x2≤≤ , −+ +≤ 33 4 4x x x x 3 3 . Solution. Applying the Cauchy – Buniakowski Inequality we get ( ) () () ⎡⎤ −+ + = − + + ≤ + − ++ = ⎣⎦ 33 3 3 33 11 4x x x x 2 8x 2x 2 x x 2 4 8x 2x x x 22 () ()() 3 22 2 2 22 4 4 4 44 2x 9x 9x 16 6 .2x 9 x 33 23 222 ⎛⎞ +− +− ⎜⎟ =−= = ⎜⎟ ⎝⎠ . 17. () x ∈  , + −≤2sinx 15 10 2cosx 6. Solution. Applying the Cauchy – Buniakowski Inequality we get () ()() ⎡⎤ +− ≤+ +− = − +≤ ⎢⎥ ⎣⎦ 2 2 2 sin x 15 10 2 cosx 1 5 2sin x 3 2 2 cosx 6 6 2 cosx 1 6 . 18. () 222 x, y,z 0,x y z 1,n + >++=∈ , ( ) + + ++≥ −−− 2n 2n 2n 2n 2n 1 2n 1 xyz 1x 1y 1z 2n . Solution. We set () () () 2n ft t1 t , t 0,1=− ∈ . It is easy to show that () ( ) () () () () + ⎛⎞ =− + = ⇔= ⇒ ≤ ⇒ − ≤ ⎜⎟ + +++ ⎝⎠ 2n 1 2n 2n 2n 2n 11 2n f' t 1 2n 1 t , f' t 0 t f t f t 1 t 2n 1 2n 1 2n 1 2n 1 Since 0x,y,z1<<, thus () () 2n 2n 2n 1 2n 1 1 , 2n x1 x + + ≥ − () ( ) 2n 2n 2n 1 2n 1 1 , 2n y1 y + + ≥ − () () 2n 2n 2n 1 2n 1 1 2n z1 z ++ ≥ − . 100 INEQUALITY PROBLEMS ***** Cao Minh Quang 05/11/2006 Most of these problems were collected. 2 444 22 22 55 5 555 ⎡⎤ ⎛⎞⎛⎞ ++++ ++++ ⎢⎥ ⎜⎟⎜⎟ ++ ⎡⎤ ≤+== ⎢⎥ ⎜⎟⎜⎟ ⎢⎥ ⎢⎥ ⎣⎦ ⎜⎟⎜⎟ ⎜⎟⎜⎟ ⎢⎥ ⎝⎠⎝⎠ ⎣⎦ . Third solution. Applying the AM – GM Inequality we get 1 ab2ab ab 4 + ≥⇒≤ . Therefore, () 22 2 22. bbb ccc 33 333 44 4 4 4 ⎡⎤ ⎛⎞⎛⎞⎛⎞ +++ +++ +++ ⎢⎥ ⎜⎟⎜⎟⎜⎟ ≤++= ⎢⎥ ⎜⎟⎜⎟⎜⎟ ⎢⎥ ⎜⎟⎜⎟⎜⎟ ⎢⎥ ⎝⎠⎝⎠⎝⎠ ⎣⎦ . Third solution. Applying the AM – GM Inequality we get ()()() ⎛⎞ ++ +++ + + ≥ ⎡⎤ ⎜⎟ ⎣⎦ +++ ⎝⎠ 111 a1