Tổng hợp 567 Bất đẳng thức hay và khó có lời giải chi tiết - Nguyễn Duy Tùng

Nguyen Duy Tung 567 Nice And Hard Inequalities ====================================== ============================================== 1 This product is created for educational purpose. Please don't use use it for any commecial purpose unless you got the right of the author. Please contact Email:[email protected] 2 1. a) if a, b, c are positive real numbers, then r r r a2 + 1 b2 + 1 c2 + 1 a b c + + ≥ + + . b c a b2 + 1 c2 + 1 a2 + 1 b)Let a, b, c, d be positive real numbers.Prove that a2 − bd b2 − ca c2 − db d2 − ac + + + ≥ 0. b + 2c + d c + 2d + a d + 2a + b a + 2b + c Solution: a)By Cauchy-Schwarz's inequality, We have:
p
a2 + b2 (a2 + 1) (b2 + 1) ≥ a2 + b2 (ab + 1)

= ab a2 + b2 + a2 + b2 ≥ ab a2 + b2 + 2 Xa Xb X a2 + b2 + = ⇒ b a ab r r 2 2 2 X X a +b +2 a + 1 X b2 + 1 p = + ≥ b2 + 1 a2 + 1 (a2 + 1) (b2 + 1) By Chebyshev's inequality, We have X a2 b2 = Therefore X X X a2 X X a2 + 1 a2 a2 b2 + ≥ + = . b2 + 1 b2 (b2 + 1) b2 + 1 b2 (b2 + 1) b2 + 1  a Xb a2 + + 2 b b a b ! r r X a2 + 1 X b2 + 1 X a2 + 1 ≥1+2 + + 2 2 b +1 a +1 b2 + 1 !2 r X a2 + 1 . = 1+ b2 + 1  1+ X a 2 Therefore a b c + + ≥ b c a =1+2 r X a2 + 1 + b2 + 1 r b2 + 1 + c2 + 1 r c2 + 1 a2 + 1 as require. b)Notice that 2a2 + b2 + d2 + 2c(b + d) 2(a2 − bd) +b+d= b + 2c + d b + 2c + d 2 2 (a − b) + (a − d) + 2(a + c)(b + d) = (1) b + 2c + d And similarly, 2(c2 − db) (c − d)2 + (c − b)2 + 2(a + c)(b + d) +b+d= (2) d + 2a + b d + 2a + b Using Cauchy-Schwarz's inequality,we get (a − d)2 (c − d)2 [(a − b)2 + (c − d)2 ] + ≥ (3) b + 2c + d d + 2a + b (b + 2c + d) + (d + 2a + b) 3 (a − d)2 (c − b)2 [(a − d)2 + (c − b)2 ]2 + ≥ (4) b + 2c + d d + 2a + b (b + 2c + d) + (d + 2a + b) 2(a + c)(b + d) 2(a + c)(b + d) 8(a + c)(b + d) + ≥ (5) b + 2c + d d + 2a + b (b + 2c + d) + (d + 2a + b) From (1),(2),(3),(4) and (5), we get 2( c2 − db (a + c − b − d)2 + 4(a + c)(b + d) a2 − bd + )+b+d≥ = a + b + c + d. b + 2c + d d + 2a + b a+b+c+d or a2 − bd c2 − db a+c−b−d + ≥ b + 2c + d d + 2a + b 2 In the same manner,we can also show that b2 − ca d2 − ac b+d−a−c + ≥ c + 2d + a a + 2b + c 2 and by adding these two inequalities,we get the desired result. Enquality holds if and only if a = c and b = d. 2, Let a, b, c be positive real numbers such that a+b+c=1 Prove that the following inequality holds bc ca 3 ab + + ≤ 1 − c2 1 − a2 1 − b2 8 Solution: From the given condition The inequality is equivalent to X 4ab 3 ≤ a2 + b2 + 2(ab + bc + ca) 2 but from Cauhy Shwarz inequality X ≤ 4ab a2 + b2 + 2(ab + bc + ca) X ab ab + 2 2 a + ab + bc + ca b + ab + bc + ca X X ab ab = + (a + b)(a + c) (b + c)(a + b) X a(b + c)2 = (a + b)(b + c)(c + a) Thus We need prove that 3(a + b)(b + c)(c + a) ≥ 2 X which reduces to the obvious inequality X ab(a + b) ≥ 6abc The Solution is completed.with equality if and only if a=b=c= 4 1 3 a(b + c)2  Or We can use the fact that X a2 + b2 X 4ab 4ab ≤ + 2(ab + bc + ca) (2ab + 2ac) + (2ab + 2bc) X X ab ab + ≤ 2a(b + c) 2b(a + c)   X 1 b a = + 2 b+c a+c   1X b c 3 = + = 2 b+c b+c 2 3, Let a, b, c be the positive real numbers. Prove that p 4. 3 (a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab) ab2 + bc2 + ca2 ≥ 1+ (ab + bc + ca)(a + b + c) (a + b + c)2 Solution: Multiplying both sides of the above inequality with (a + b + c)2 it's equivalent to prove that (a + b + c)(ab2 + bc2 + ca2 ) ab + bc + ca (a + b + c)2 + p ≥ 4. 3 (a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab) We have (a + b + c)2 + (a + b + c)(ab2 + bc2 + ca2 ) X (a2 + ab + bc)(c + a)(c + b) = ab + bc + ca ab + bc + ca By using AM-GM inequality We get X (a2 + ab + bc)(c + a)(c + b) ab + bc + ca p 3 ≥ 3. (a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab)[(a + b)(b + c)(c + a)]2 ab + bc + ca Since it's suffices to show that √ p √ 3. 3 (a + b)(b + c)(c + a) ≥ 2. ab + bc + ca which is clearly true by AM-GM inequality again. The Solution is completed. Equality holds for a = b = c 4, Let a0 , a1 , . . . , an be positive real numbers such that ak+1 − ak ≥ 1 for all k = 0, 1, . . . , n − 1. Prove that          1 1 1 1 1 1 1+ 1+ ··· 1 + ≤ 1+ 1+ ··· 1 + a0 a1 − a0 an − a0 a0 a1 an Solution: We will prove it by induction. For n = 1 We need to check that      1 1 1 1 1+ 1+ ≤ 1+ 1+ a0 a1 − a0 a0 a1 which is equivalent to a0 (a1 − a0 − 1) ≥ 0, which is true by given condition. Let          1 1 1 1 1 1 1+ 1+ ··· 1 + ≤ 1+ 1+ ··· 1 + a0 a1 − a0 ak − a0 a0 a1 ak 5 it remains to prove that:     1 1 1 1+ 1+ ··· 1 + ≤ a0 a1 − a0 ak+1 − a0      1 1 1 ≤ 1+ 1+ ··· 1 + a0 a1 ak+1 By our hypothesis     1 1 1 1+ ··· 1 + ≥ a0 a1 ak+1       1 1 1 1 ≥ 1+ 1+ 1+ ··· 1 + ak+1 a0 a1 − a0 ak − a0  1+ id est, it remains to prove that:       1 1 1 1 1+ 1+ ··· 1 + ≥ 1+ ak+1 a0 a1 − a0 ak − a0     1 1 1 ≥1+ 1+ ··· 1 + a0 a1 − a0 ak+1 − a0 But       1 1 1 1+ 1+ ··· 1 + ≥ ak+1 a0 a1 − a0 ak − a0     1 1 1 ≥1+ 1+ ··· 1 + ⇔ a0 a1 − a0 ak+1 − a0     1 1 1 1 + 1+ ··· 1 + ≥ ⇔ ak+1 ak+1 a0 a1 − a0 ak − a0     1 1 1 ≥ 1+ ··· 1 + ⇔ (ak+1 − a0 )a0 a1 − a0 ak − a0     1 1 1 ⇔1≥ 1+ ··· 1 + ak+1 − a0 a1 − a0 ak − a0 1+ 1 But by our conditions We obtain:  1 1+ ak+1 − a0  1 ≤ 1+ k    1 1 ··· 1 + ≤ a1 − a0 ak − a0    1 1 ··· 1 + = 1. 1 k−1 Thus, the inequality is proven. 5, Given a, b, c > 0. Prove that X r 3 √ 3 a2 + bc abc ≥ 9. b2 + c2 (a + b + c) Solution : This ineq is equivalent to: a2 + bc X q 3 ≥ 2 abc(a2 + bc) (b2 + c2 ) 9 3 (a + b + c) By AM-GM ineq , We have a2 + bc q 3 2 abc(a2 + bc) (b2 + c2 ) 6 = = p 3 a2 + bc (a2 + bc)c(a2 + bc)b(b2 + c2 )a ≥ 3(a2 + bc) P 2 a b sym Similarly, this ineq is true if We prove that: 3(a2 + b2 + c2 + ab + bc + ca) 9 P 2 ≥ 3 a b (a + b + c) sym a3 + b3 + c3 + 3abc ≥ X a2 b sym Which is true by Schur ineq. Equality holds when a = b = c 6, Let a, b, c be nonnegative real numbers such that ab + bc + ca > 0. Prove that 1 1 1 2 + + ≥ . 2a2 + bc 2b2 + ca 2c2 + ab ab + bc + ca The inequality is equivalent to X ab + bc + ca 2a2 + bc ≥ 2, (1) or X a(b + c) X bc + ≥ 2.(2) 2a2 + bc bc + 2a2 Using the Cauchy-Schwarz inequality, We have P 2 X ( bc) bc ≥P = 1.(3) bc + 2a2 bc(bc + 2a2 ) Therefore, it suffices to prove that X a(b + c) ≥ 1.(4) 2a2 + bc Since a(b + c) a(b + c) ≥ 2 2a + bc 2(a2 + bc) it is enough to check that X a(b + c) a2 + bc ≥ 2, (5) which is a known result. Remark: 2ca + bc 2bc + ca 4c + 2 ≥ . 2 2a + bc 2b + ca a+b+c 7, Let a, b, c be non negative real numbers such that ab + bc + ca > 0. Prove that 2a2 1 1 1 1 12 + 2 + 2 + ≥ . + bc 2b + ca 2c + ab ab + bc + ca (a + b + c)2 Solution: 1) We can prove this inequality using the following auxiliary result if 0 ≤ a ≤ min{a, b}, then 1 1 4 + ≥ . 2a2 + bc 2b2 + ca (a + b)(a + b + c) 7 in fact, this is used to replaced for "no two of which are zero", so that the fractions 1 1 1 1 , , , 2a2 + bc 2b2 + ca 2c2 + ab ab + bc + ca have meanings. Besides, the iaker also works for it: 2a2 1 1 1 2(ab + bc + ca) + 2 + 2 ≥P 2 2 + bc 2b + ca 2c + ab a b + abc(a + b + c) But our Solution for both of them is expand Let a, b, c be non negative real numbers such that ab + bc + ca > 0. Prove that 1 1 1 1 12 . + + + ≥ 2a2 + bc 2b2 + ca 2c2 + ab ab + bc + ca (a + b + c)2 2) Consider by AM-GM inequality, We have
2 a2 + ab + b2 (a + b + c)

= (2b + a) 2a2 + bc + (2a + b) 2b2 + ca p ≥ 2 (2a + b)(2b + a) (2a2 + bc) (2b2 + ca). And by AM-GM inequality, We have c2 (2a + b) c2 (2b + a) + 2a2 + bc 2b2 + ca s c4 (2a + b)(2b + a) ≥2 (2a2 + bc) (2b2 + ca) 2c2 (2a + b)(2b + a) + ab + b2 ) (a + b + c)   4c2 6abc c = + a + b + c a + b + c a2 + ab + b2 ≥ (a2 X 2c2 a + bc2 + 2ab2 + b2 c 2a2 + bc  c2 (2a + b) c2 (2b + a) + 2a2 + bc 2b2 + ca   X  4c2 6abc c ≥ + a + b + c a + b + c a2 + ab + b2
X  4 a2 + b2 + c2 6abc c = + a+b+c a+b+c a2 + ab + b2
  4 a2 + b2 + c2 6abc (a + b + c)2 P ≥ + a+b+c a+b+c c (a2 + ab + b2 )
4 a2 + b2 + c2 6abc = + ab + c ab + bc + ca X 2a2 b + 2ab2 + 2b2 c + 2bc2 + 2c2 a + 2ca2 ⇒ 2a2 + bc = X 8 X X 2c2 a + bc2 + 2ab2 + b2 c (b + c) + 2a2 + bc
X 4 a2 + b2 + c2 6abc ≥ (b + c) + + a+b+c ab + bc + ca

P 2 8 a2 + b2 + c2 + ab + bc + ca 2 a b + ab2 = − a+b+c ab + bc + ca X 1 1 + ⇒ 2a2 + bc ab + bc + ca
4 a2 + b2 + c2 + ab + bc + ca P ≥ (a + b + c) ( (a2 b + ab2 )) = ≥ <=> X (a + b)(a + c) 2a2 + bc From + 12 . (a + b + c)2 X a2 + bc X 2a2 + 2bc 2a2 We get + bc 2a2 + bc −3= −2≥ 12(ab + bc + ca) (a + b + c)2 bc ≥1 + bc 2a2 X a2 + bc −2≥0 2a2 + bc Now, We will prove the stronger X (a + b)(a + c) 2a2 + bc ≥ 12(ab + bc + ca) (a + b + c)2 From cauchy-scharzt, We have X (a + b)(a + c) 2a2 + bc X = (a+b)(b+c)(c+a)( (2a2 3(a + b)(b + c)(c + a) 1 ≥ + bc)(b + c) ab(a + b) + bc(b + c) + ca(c + a) Finally, We only need to prove that 4(ab + bc + ca) (a + b)(b + c)(c + a) ≥ ab(a + b) + bc(b + c) + ca(c + a) (a + b + c)2 (a + b + c)2 4[ab(a + b) + bc(b + c) + ca(c + a) 8abc ≥ =4− ab + bc + ca (a + b)(b + c)(c + a) (a + b)(b + c)(c + a) 8abc a2 + b2 + c2 + ≥2 ab + bc + ca (a + b)(b + c)(c + a) which is old problem. Our Solution are completed equality occur if and if only a = b = c, a = b, c = 0 or any cyclic permution. 8, Let a, b, c be positive real numbers such that 16(a + b + c) ≥ X 1 a + 1 b + 1c . Prove that 1 8 h i3 ≤ . p 9 a + b + 2(a + c) Solution: This problem is rather easy. Using the AM-GM inequality, We have: p a + b + 2(c + a) = a + b + r c+a + 2 9 r r c+a 3 (a + b)(c + a) ≥3 . 2 2 So that: X X 1 2 . ≤ h i 3 p 27(a + b)(c + a) a + b + 2(c + a) Thus, it's enough to check that: X 1 ≤ 4 ⇐⇒ 6(a + b)(b + c)(c + a) ≥ a + b + c, 3(a + b)(c + a) which is true since 9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca) and 16abc(a + b + c) ≥ ab + bc + ca ⇒ 16(ab + bc + ca)2 3 ≥ ab + bc + ca ⇐⇒ ab + bc + ca ≥ . 3 16 The Solution is completed. Equality holds if and only if a = b = c = 41 . 9, Let x, y, z be positive real numbers such that xyz = 1. Prove that x3 + 1 p x4 + y + z √ z3 + 1 +p ≥ 2 xy + yz + zx. y4 + z + x z4 + x + y +p y3 + 1 Solution: Using the AM-GM inequality, We have p p 2 (x4 + y + z)(xy + yz + zx) = 2 [x4 + xyz(y + z)](xy + yz + zx) p = 2 (x3 + y 2 z + yz 2 )(x2 y + x2 z + xyz) ≤ (x3 + y 2 z + yz 2 ) + (x2 y + x2 z + xyz) = (x + y + z)(x2 + yz) = it follows that x3 + 1 p x4 + y + z ≥ (x + y + z)(x3 + 1) . x √ 2x xy + yz + zx . x+y+z Adding this and it analogous inequalities, the result follows. 10, Let a, b, c be nonnegative real numbers satisfying a + b + c = √ (a2 − b2 )(b2 − c2 )(c2 − a2 ) ≤ 5 √ 5. Prove that Solution: For this one, We can assume WLOG that c ≥ b ≥ a so that We have P = (a2 − b2 )(b2 − c2 )(c2 − a2 ) = (c2 − b2 )(c2 − a2 )(b2 − a2 ) ≤ b2 c2 (c2 − b2 ). Also note that have √ 5 = a + b + c ≥ b + c since a ≥ 0. Now, using the AM-GM inequality We (c + b) · ! !2 √ 5 + 1 b · (c − b) 2 (√ )5 √ 5(b + c) ≤ (c + b) ≤ 5; 5 ! !2 √ 5 −1 ·c · 2 √ So get P ≤ 5. And hence We are done. Equality holds if and only if (a, b, c) =  √ that We √ 5 5 2 + 1; 2 − 1; 0 and all its cyclic permutations. 2 10 11, Let a, b, c > 0 and a + b + c = 3. Prove that 1 1 3 1 + + ≤ 2 2 2 2 2 2 3+a +b 3+b +c 3+c +a 5 Solution: We have: 1 1 1 3 + + ≤ 3 + a2 + b2 3 + b2 + c2 3 + c2 + a2 5 3 3 9 3 + + ≤ 3 + a2 + b2 3 + b2 + c2 3 + c2 + a 2 5 X a2 + b2 6 ≥ 3 + a2 + b2 5 Using Cauchy-Schwarz's inequality:  X X Xp a2 + b2 2 2 ( 3 + a + b ) ≥ ( a2 + b2 )2 3 + a2 + b2 <=> That means We have to prove Xp 6 X (3 + a2 + b2 )) a2 + b2 )2 ≥ ( ( 5 X Xp 54 12 X 2 (a2 + b2 ) + 2 (a2 + b2 )(a2 + c2 ) ≥ + a 5 5 X X X 8 a2 + 10 ab ≥ 54 <=> 5(a + b + c)2 + 3 a2 ≥ 54 it is true with a + b + c = 3. 12, Given a, b, c > 0 such that ab + bc + ca = 1. Prove that 1 1 1 + + ≥1 4a2 − bc + 1 4b2 − ca + 1 4c2 − ab + 1 Solution: in fact, the sharper inequality holds 1 1 1 3 + + ≥ . 4a2 − bc + 1 4b2 − ca + 1 4c2 − ab + 1 2 The inequality is equivalent to 1 1 1 3 + + ≥ . a(4a + b + c) b(4b + c + a) c(4c + a + b) 2 Using the Cauchy-Schwarz inequality, We have X 1 a(4a + b + c)  X 4a + b + c a  ≥ X 2 1 1 = 2 2 2. a a b c Therefore, it suffices to prove that 2 4a + b + c 4b + c + a 4c + a + b + + . ≥ 3a2 b2 c2 a b c Since X 4a + b + c a X a+b+c a a+b+c =9+ , abc = 3+ 11  =9+ (a + b + c)(ab + bc + ca) abc this inequality can be written as 9a2 b2 c2 + abc(a + b + c) ≤ which is true because a2 b2 c2 ≤  ab + bc + ca 3 3 = 2 , 3 1 , 27 and (ab + bc + ca)2 1 = . 3 3 13, Given a, b, c ≥ 0 such that ab + bc + ca = 1. Prove that abc(a + b + c) ≤ 1 1 1 + + ≥1 4a2 − bc + 2 4b2 − ca + 2 4c2 − ab + 2 Solution: Notice that the case abc = 0 is trivial so let us consider now that abc > 0. Using the AM-GM inequality, We have [c(2a + b) + b(2a + c)]2 4bc 1 (ab + bc + ca)2 = . = bc bc 4a2 − bc + 2(ab + bc + ca) = (2a + b)(2a + c) ≤ it follows that 1 ≥ bc. 4a2 − bc + 2 Adding this and its analogous inequalities, We get the desired result. 14, Given a, b, c are positive real numbers. Prove that 1 1 1 1 1 1 9 ( + + )( + + )≥ . a b c 1+a 1+b 1+c 1 + abc Solution: The original inequality is equivalent to  abc + 1 abc + 1 abc + 1 + + a b c or X 1 + a2 c cyc a !  1 1 1 + + a+1 b+1 c+1 1 1 1 + + a+1 b+1 c+1  ≥9  ≥9 By Cauchy Schwarz ineq and AM-GM ineq, X 1 + a2 c cyc and a ≥ X c(1 + a)2 cyc a(1 + c) p ≥ 3 3 (1 + a)(1 + b)(1 + c) 1 1 1 3 + + ≥ p 3 a+1 b+1 c+1 (1 + a)(1 + b)(1 + c) Multiplying these two inequalities, the conclusion follows. Equality holds if and only if a = b = c = 1. 15. Given a, b, c are positive real numbers. Prove that: p p p 3p a(b + 1) + b(c + 1) + c(a + 1) ≤ (a + 1)(b + 1)(c + 1) 2 12 Solution: Case1.if a + b + c + ab + bc + ca ≤ 3abc + 3 <=> 4(ab + bc + ca + a + b + c) ≤ 3(a + 1)(b + 1)(c + 1) Using Cauchy-Schawrz's inequality ,We have: p p p 9(a + 1)(b + 1)(c + 1) ( a(b + 1) + b(c + 1) + c(a + 1))2 ≤ 3(ab + bc + ca + a + b + c) ≤ 4 The inequality is true. Case2. ifa + b + c + ab + bc + ca ≤ 3abc + 3. <=> 9(a + 1)(b + 1)(c + 1) ≥ 2(a + b + c + ab + bc + ca) + 3abc + 3 4 By AM-GM's inequality : Xp X 2 ab(b + 1)(c + 1) ≤ [ab(c + 1) + (b + 1)] = a + b + c + ab + bc + ca + 3abc + 3 => ab + bc + ca + a + b + c + 2 Xp ab(b + 1)(c + 1) ≤ 9 4(a + 1)(b + 1)(c + 1) p p p 3p (a + 1)(b + 1)(c + 1)]2 => ( a(b + 1) + b(c + 1) + c(a + 1))2 ≤ [ 2 => Q.E.D Enquality holds when a = b = c = 1. 16, Given a, b, c are positive real numbers. Prove that: 1 1 1 10 + 2 + 2 ≥ a2 + b2 b + c2 c + a2 (a + b + c)2 Solution: Assume c = min{a, b, c}. Then a2 1 1 2 + 2 ≥ ⇐⇒ (ab − c2 )(a − b)2 ≥ 0 2 2 +c b +c ab + c2 And by Cauchy-schwarz 2 2  2 ((a + b ) + 8(ab + c )) 1 2 + a2 + b2 ab + c2  ≥ 25 Hence We need only to prove: 5(a + b + c)2 ≥ 2((a2 + b2 ) + 8(ab + c2 )) ⇐⇒ 3(a − b)2 + c(10b + 10a − 11c) ≥ 0 Equality for a = b, c = 0 or permutations. 17, Let a, b and c are non-negative numbers such that ab + ac + bc 6= 0. Prove that b2 (a + c)2 c2 (a + b)2 a2 (b + c)2 + + ≤ a2 + b2 + c2 a2 + 3bc b2 + 3ac c2 + 3ab Solution: By Cauchy-Schwarz ineq , We have 2 3 3 a2 (b + c) a2 (b + c) a2 (b + c) = 2 = 2 2 a + bc (a + bc)(b + c) b(a + c2 ) + c(a2 + b2 ) ≤ a2 (b + c) b2 c2 a2 (b + c) b c ( 2 + ) = ( 2 + 2 ) 2 2 2 2 4 b(a + c ) c(a + b ) 4 a +c a + b2 Similarly, We have LHS ≤ X a2 (b + c)( X c(a2 (b + c) + b2 (c + a)) b c + ) = a2 + c2 a2 + b2 a2 + b2 13 = a2 + b2 + c2 + X abc(a + b) a2 b2 + which is true by AM-GM ineq ≤ a2 + b2 + c2 + X abc(a + b) a2 + b2 ≤ a2 + b2 + c2 + ab + bc + ca The original inequality can be written as X (a + b)2 (a + c)2 a2 + bc ≤ 8 (a + b + c)2 . 3 Since (a + b)(a + c) = (a2 + bc) + a(b + c) We have (a2 + bc)2 + 2a(b + c)(a2 + bc) + a2 (b + c)2 (a + b)2 (a + c)2 = a2 + bc a2 + bc a2 (b + c)2 = a2 + bc + 2a(b + c) + 2 , a + bc and thus the above inequality is equivalent to X a2 (b + c)2 a2 + bc or ≤ X a2 (b + c)2 a2 + bc X X 8 (a + b + c)2 − a2 − 5 ab, 3 ≤ 5(a2 + b2 + c2 ) + ab + bc + ca . 3 Since 5(a2 + b2 + c2 ) + ab + bc + ca ≥ a2 + b2 + c2 + ab + bc + ca 3 it is enough show that X a2 (b + c)2 a2 + bc ≤ a2 + b2 + c2 + ab + bc + ca. Q.E.D 18, Given a1 ≥ a2 ≥ . . . ≥ an ≥ 0, b1 ≥ b2 ≥ . . . ≥ bn ≥ 0 n X ai = 1 = i=1 Find the maxmium of n X bi i=1 n X (ai − bi )2 i=1 WSolution:ithout loss of generality, assume that a1 ≥ b1 Notice that for a ≥ x ≥ 0, b, y ≥ 0 We have (a − x)2 + (b − y)2 − (a + b − x)2 − y 2 = −2b(a − x + y) ≤ 0. According to this inequality, We have (a1 − b1 )2 + (a2 − b2 )2 ≤ (a1 + a2 − b1 )2 + b22 , (a1 + a2 − b1 )2 + (a3 − b3 )2 ≤ (a1 + a2 + a3 − b1 )2 + b23 , · · · · · · 14 (a1 + a2 + · · · + an−1 − b1 )2 + (an − bn )2 ≤ (a1 + a2 + · · · + an − b1 )2 + b2n . Adding these inequalities, We get n X (ai − bi )2 ≤ (1 − b1 )2 + b22 + b23 + · · · + b2n i=1 ≤ (1 − b1 )2 + b1 (b2 + b3 + · · · + bn ) = (1 − b1 )2 + b1 (1 − b1 ) = 1 − b1 ≤ 1 − 1 . n Equality holds for example when a1 = 1, a2 = a3 = · · · = an = 0 and b1 = b2 = · · · = bn = 1 n 19, Given a, b, c ≥ 0 such that a2 + b2 + c2 = 1 Prove that 1 − ab 1 − bc 1 − ca 1 + + ≥ 7 − 3ac 7 − 3ba 7 − 3cb 3 Solution: First, We will show that 1 1 1 1 + + ≤ . 7 − 3ab 7 − 3bc 7 − 3ca 2 Using the Cauchy-Schwarz inequality, We have   1 1 1 1 = ≤ +1 . 7 − 3ab 3(1 − ab) + 4 9 3(1 − ab) it follows that 1 1 X 1 1 ≤ + , 7 − 3ab 27 1 − ab 3 and thus, it is enough to show that X 1 1 9 1 + + ≤ , 1 − ab 1 − bc 1 − ca 2 which is Vasc's inequality. Now, We write the original inequality as 3 − 3ab 3 − 3bc 3 − 3ca + + ≥ 1, 7 − 3ac 7 − 3ba 7 − 3cb or 7 − 3bc 7 − 3ca 7 − 3ab + + ≥1+4 7 − 3ac 7 − 3ba 7 − 3cb Since  4  1 1 1 + + 7 − 3ab 7 − 3bc 7 − 3ca 1 1 1 + + 7 − 3ab 7 − 3bc 7 − 3ca 15  ≤2  . it is enough to show that 7 − 3ab 7 − 3bc 7 − 3ca + + ≥ 3, 7 − 3ac 7 − 3ba 7 − 3cb which is true according to the AM-GM inequality. 21, Let a, b, c ≥ 0 such that a+b+c>0 and b + c ≥ 2a For x, y, z > 0 such that xyz = 1 Prove that the following inequality holds 1 1 3 1 + + ≥ a + x2 (by + cz) a + y 2 (bz + cx) a + z 2 (bx + cy) a+b+c Solution: Setting u= 1 1 ,v = x y and w= 1 z and using the condition uvw = 1 the inequality can be rewritten as X X u u2 3 = > . au + cv + bw au2 + cuv + bwu a+b+c Applying Cauchy, it suffices to prove 2 (u + v + w) 3 P > a u2 + (b + c) uv a+b+c X  1 · (b + c − 2a) (x − y)2 > 0, 2 which is obvious due to the condition for P a, b, c 22, Given x, y, z > 0 such that xyz = 1 16 Prove that 1 (1 + x2 )(1 + x7 ) + 1 (1 + y 2 )(1 + y7 ) + 1 (1 + z 2 )(1 + z7) ≥ Solution: First We prove this ineq easy 1 (1 + x2 )(1 + x7 ) ≥ 3 9 4(x9 + x 2 + 1) And this ineq became: 1 x9 9 2 +x +1 + 1 y9 9 2 +y +1 + 1 z9 9 + z2 + 1 ≥1 with xyz = 1 it's an old result 23, Let a, b, c be positive real numbers such that 3(a2 + b2 + c2 ) + ab + bc + ca = 12 Prove that √ 3 a b c ≤√ . +√ +√ c + a a+b b+c 2 Solution: Let A = a2 + b2 + c2 , B = ab + bc + ca X X 3 X 2 X  3 a + ab = 9. 2A + B = 2 a2 + ab ≤ 4 By Cauchy Schwarz inequality, We have X X√ r a a √ = a a+b a+b rX √ a ≤ a+b+c . a+b By Cauchy Schwarz inequality again, We have X X b b2 = a+b b(a + b) (a + b + c)2 ≥ P b(a + b) = A + 2B A+B X b a A + 2B 2A + B =3− ≤3− = a+b a+b A+B A+B hence, it suffices to prove that X (a + b + c) · 2A + B 9 ≤ A+B 2 17 3 4 Consider √ (a + b + c) 2A + B p = (A + 2B) (2A + B) (A + 2B) + (2A + B) 2 3 = (A + B) 2 2A + B 3√ 9 ⇒ (a + b + c) · ≤ 2A + B ≤ A+B 2 2 ≤ as require. By AM-GM ineq easy to see that 3 ≤ a2 + b2 + c2 ≤ 4 By Cauchy-Schwarz ineq, We have √ X X a a a+c 2 p ) ≤ (a2 + b2 + c2 + ab + bc + ca)( ) LHS = ( (a + b)(a + c) (a + b)(a + c) Using the familiar ineq 9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca) We have X 2(ab + bc + ca) 9 a = ≤ (a + b)(a + c) (a + b)(b + c)(c + a) 4(a + b + c) And We need to prove that 9 6 − (a2 + b2 + c2 ) 9(a2 + b2 + c2 + ab + bc + ca) ≤ ⇔p ≤1 4(a + b + c) 2 24 − 5(a2 + b2 + c2 ) ⇔ (6 − (a2 + b2 + c2 ))2 ≤ 24 − 5(a2 + b2 + c2 ) ⇔ (3 − (a2 + b2 + c2 ))(4 − (a2 + b2 + c2 )) ≤ 0 Which is true We are done equality holds when a=b=c=1 24. Given a, b, c ≥ 0 Prove that X 8(a + b + c)2 1 ≤ (a2 + bc)(b + c)2 3(a + b)2 (b + c)2 (c + a)2 Solution: in fact, the sharper and nicer inequality holds: b2 (c + a)2 c2 (a + b)2 a2 (b + c)2 + + ≤ a2 + b2 + c2 + ab + bc + ca. a2 + bc b2 + ca c2 + ab a2 (b + c)2 b2 (c + a)2 c2 (a + b)2 + + ≤ a2 + b2 + c2 + ab + bc + ca a2 + bc b2 + ca c2 + ab 18 25. Given a, b, c ≥ 0 such that ab + bc + ca = 1 Prove that 8 2 5a 1 1 1 9 + 8 2 + 8 2 ≥ 4 + bc b + ca c + ab 5 5 Assume WLOG a≥b≥c this ineq 8 2 5a 1 5 1 5 1 − + 8 2 − + 8 2 −1≥0 8 + bc 8 5 b + ca 5 c + ab 8 − 8a2 − 5bc 8 − 8b2 − 5ca 1 − 58 c2 − ab ≥0 + + 8a2 + 5bc 8b2 + 5ca c2 + 85 ab 8a(b + c − a) + 3bc 8b(a + c − b) + 5ac c(a + b − 85 c) + + ≥0 8a2 + 5bc 8b2 + 5ca c2 + 85 ab Notice that We only need to prove this ineq when a≥b+c by the way We need to prove that 8b2 8b 8a ≥ 2 + 5ca 8a + 5bc (a − b)(8ab − 5ac − 5bc) ≥ 0 Easy to see that: if a≥b+c then 8ab = 5ab + 3ab ≥ 5ac + 6bc ≥ 5ac + 5ac So this ineq is true, We have q.d.e , equality hold when (a, b, c) = (1, 1, 0) 26, Give a, b, c ≥ 0 Prove that: b2 a b c a+b+c abc(a + b + c) + 2 + 2 ≥ + 3 2 2 2 +c a +c a +b ab + bc + ca (a + b3 + c3 )(ab + bc + ca) X b2 X a a2 = 2 2 +c ab + c2 a (a + b + c)2 , ≥P (ab2 + c2 a) it suffices to prove that P 1 a+b+c abc ≥ + , (ab2 + c2 a) ab + bc + ca (ab + bc + ca) (a3 + b3 + c3 ) 19 because P = a+b+c 1 − 2 2 (ab + c a) ab + bc + ca 3abc P , (ab + bc + ca) (ab2 + ca2 ) it suffices to prove that
X
3 a3 + b3 + c3 ≥ ab2 + c2 a , which is true because
X
2 a3 + b3 + c3 ≥ ab2 + c2 a . Remark: a b c a+b+c 3abc(a + b + c) + 2 + 2 ≥ + . b2 + c2 c + a2 a + b2 ab + bc + ca 2(a3 + b3 + c3 )(ab + bc + ca) Give a, b, c ≥ 0 Prove that a2 1 1 3 81a2 b2 c2 1 + 2 + 2 ≥ + 2 + bc b + ca c + ab ab + bc + ca 2(a + b2 + c2 )4 Equality occur if and if only a = b = c, a = b, c = 0 or any cyclic permution. it is true because (1)
3 a2 + b2 + c2 1 1 1 + + ≥ 3 a2 + bc b2 + ca c2 + ab a b + ab3 + b3 c + bc3 + c3 a + ca3 and
3 a2 + b2 + c2 3 81a2 b2 c2 (2) 3 ≥ + . a b + ab3 + b3 c + bc3 + c3 a + ca3 ab + bc + ca 2(a2 + b2 + c2 )4 Because P P = (a3 b a2 1 − + ab3 ) ab + bc + ca abc(a + b + c) P , (ab + bc + ca) ( (a3 b + ab3 )) it suffices to prove that 2(a + b + c) a2 + b2 + c2
4 ≥ 27abc(ab + bc + ca) X which is true because
(a) (a + b + c) a2 + b2 + c2 ≥ 9abc, (b) a2 + b2 + c2 ≥ ab + bc + ca, X
2
(c) 2 a2 + b2 + c2 ≥ 3 a3 b + ab3 , which (c) 20 a3 b + ab3
 ,