Several proofs and generalizations of a fractionalinequality

Several proofs and generalizations of a fractional inequality with constraints Fuhua Wei and Shanhe Wu ∗ Department of Mathematics and Computer Science, Longyan University, Longyan, Fujian 364012, P R China E-mail: wushanhe@yahoo.com.cn ∗ Corresponding Author Abstract: Ten different proofs are given for a fractional inequality with constraints Finally, two generalized forms are established by introducing exponent parameters and additive terms Keywords: fractional inequality; Cauchy-Schwarz inequality; rearrangement inequality; arithmetic-geometric means inequality; generalization 2000 Mathematics Subject Classification: 26D15 Introduction The 2nd problem given at the 36th IMO held at Toronto (Canada) in 1995 was: Problem Let a, b, c be positive real numbers with abc = Prove that 1 + + ≥ a3 (b + c) b3 (c + a) c3 (a + b) (1) In this paper, we show several different proofs and generalized forms of the inequality (1) Several proofs for the inequality (1) Proof It follows from the condition abc = that b2 c2 c2 a2 a2 b2 = , = , = + c) a(b + c) b (c + a) b(c + a) c (a + b) c(a + b) a3 (b Now, the inequality (1) is equivalent to b2 c2 c2 a2 a2 b2 + + ≥ a(b + c) b(c + a) c(a + b) By Cauchy-Schwarz inequality (see [1]), we have λ21 + λ22 + λ33 b2 c2 c2 a2 a2 b2 + + 2 λ1 λ2 λ3 ≥ (bc + ca + ab)2 Using a substitution λ21 = a(b + c), λ22 = b(c + a), λ23 = c(a + b) in the above inequality, and applying the arithmetic-geometric means inequality, we obtain b2 c2 c2 a2 a2 b2 + + ≥ (bc + ca + ab) ≥ a(b + c) b(c + a) c(a + b) 2 (abc)2 = Proof We note that the inequality (1) is equivalent to b2 c2 c2 a2 a2 b2 + + ≥ ab + ac bc + ba ca + cb Let b2 c2 c2 a2 a2 b2 + + ab + ac bc + ba ca + cb Using Cauchy-Schwarz inequality and arithmetic-geometric means inequality gives K= [(ab + ac) + (bc + ba) + (ca + cb)] K ≥ = √ ab + ac · √ √ √ bc ca ab + bc + ba · √ + ca + cb · √ ab + ac bc + ba ca + cb (bc + ca + ab)2 ≥ 3(bc + ca + ab) (bc)(ca)(ab) = 3(bc + ca + ab) Hence K ≥ 32 The desired conclusion follows Proof Note that for a > 0, a+ 1 ≥ ⇐⇒ a ≥ − a a We thus have 1 = ≥ a (b + c) 2a a (b + c) 2a 2− a2 (b + c) = ab + ac − a Similarly 1 bc + ba ≥ − , b3 (c + a) b ca + cb ≥ − + b) c c3 (a Adding the above inequalities yields 1 + + a3 (b + c) b3 (c + a) c3 (a + b) 1 1 + + − (ab + bc + ca) a b c = (ab + bc + ca) Finally, the arithmetic-geometric means inequality leads us to the required inequality ≥ Proof The inequality (1) is equivalent to b2 c2 c2 a2 a2 b2 + + ≥ a(b + c) b(c + a) c(a + b) On the other hand, we have for λ > 0, 2 √ b2 c2 + λa(b + c) ≥ λbc, a(b + c) √ c2 a2 + λb(c + a) ≥ λca, b(c + a) √ a2 b2 + λc(a + b) ≥ λab c(a + b) Adding the above inequalities yields √ ≥ (2 λ − 2λ)(ab + bc + ca) c2 a2 a2 b2 b2 c2 + + a(b + c) b(c + a) c(a + b) √ ≥ 6( λ − λ) (abc) √ = 6( λ − λ) Choosing λ = gives b2 c2 c2 a2 a2 b2 + + ≥ , a(b + c) b(c + a) c(a + b) which is the required inequality Proof We make the substitution bc = x, ca = y, ab = z, x + y + z = s Then 1 + + + c) b3 (a + c) c3 (a + b) a3 (b = = x2 y2 z2 + + y+z z+x x+y x y2 z2 + + s−x s−y s−z We consider the probability distribution sequence of random variable ξ below: p ξ= x s−x = s−x , 2s p ξ= y s−y = s−y , 2s p ξ= z s−z = s−z 2s It follows that Eξ = x s−x y s−y z s−z x+y+z · + · + · = = , s−x 2s s−y 2s s−z 2s 2s Eξ = x s−x s−x + ( − z) = 2s 2s x2 y2 z2 + + s−x s−y s−z According to D(ξ) = Eξ − (Eξ) > 0, we have 2s so x2 y2 z2 + + s−x s−y s−z ≥ , x2 y2 z2 1 3√ + + ≥ s = (x + y + z) ≥ xyz = s−x s−y s−z 2 2 Hence 1 + + ≥ a3 (b + c) b3 (a + c) c3 (a + b) Proof Let bc = x, ca = y, ab = z The inequality (1) is equivalent to y2 z2 x2 + + ≥ y+z z+x x+y By symmetry, we may assume that x ≥ y ≥ z, then x y z ≥ ≥ y+z z+x x+y Using the rearrangement inequality (see [2]) gives y2 z2 x y z x2 + + ≥z· +x· +y· , y+z z+x x+y y+z x+z x+y x2 y2 z2 x y z + + ≥y· +z· +x· , y+z z+x x+y y+z z+x x+y Adding the above inequalities yields y2 z2 x2 + + y+z z+x x+y √ ≥ x + y + z ≥ 3 xyz = 3, The required inequality follows Proof Apply the same substitution as in Proof The inequality (1) is equivalent to y2 z2 x2 + + ≥ y+z z+x x+y Since (x2 , y , z ) and ( y+z , inequality that 1 z+x , x+y ) are similarly sorted sequences, it follows from the rearrangement x2 y2 z2 + + ≥ y+z z+x x+y y2 + z2 z + x2 x2 + y + + y+z z+x x+y By the power mean inequality, we have z + x2 x2 + y y+z z+x x+y y z z x x y y2 + z2 + + ≥ + + ≥6 · · · · · y+z z+x x+y 2 2 2 2 this yields x2 y2 z2 + + ≥ y+z z+x x+y Proof Let a + b + c = t (t > 0), then 1 + + a3 (b + c) b3 (a + c) c3 (a + b) = a−2 b−2 c−2 + + b−1 + c−1 c−1 + a−1 a−1 + b−1 = a−2 b−2 c−2 + + −1 −1 t−a t−b t − c−1 Consider the following function: g(x) = x2 t−x (0 < x < t) = 3, Since 2t2 g (x) = (t − x) >0 (0 < x < t), we conclude that the function g is convex on (0, t) Using Jensen’s inequality gives a−1 + b−1 + c−1 ) 1 1 + + a b c g(a−1 ) + g(b−1 ) + g(c−1 ) ≥ 3g( = 33 1 · · a b c ≥ = Proof Consider the following function: f (x) = = √ √ bc x − ab + ac ab + ac √ + b2 c2 c2 a2 a2 b2 + + ab + ac bc + ab ac + bc √ ca x − bc + ab bc + ab + √ √ ab x − ac + bc ac + bc x2 − (ab + bc + ca) x + (ab + bc + ca) Since f (x) ≥ for x ∈ R, we have the discriminant ∆ ≤ 0, that is (ab + bc + ac) − b2 c2 c2 a2 a2 b2 + + ab + ac bc + ab ac + bc (ab + bc + ac) ≤ Thus b2 c2 c2 a2 a2 b2 3√ 3 + + ≥ (ab + bc + ac) ≥ a2 b2 c2 = , ab + ac bc + ab ac + bc 2 which leads to 1 + + ≥ a3 (b + c) b3 (c + a) c3 (a + b) Proof 10 Construct the following vectors: −→ OA = −−→ OB = a (b + c), bc a (b + c) b (c + a), ca , b (c + a) c (a + b) , , ab c (a + b) −→ −−→ We denote by θ (0 ≤ θ ≤ π) the angle of vectors OA and OB Since −→ OA = (ab + bc + ca), −−→ OB = b2 c2 c2 a2 a2 b2 + + , a (b + c) b (c + a) c (a + b) we have −→ −−→ −→ −−→ OA · OB = OA OB cos θ = (ab + bc + ca) b2 c2 c2 a2 a2 b2 + + cos θ a (b + c) b (c + a) c (a + b) ≤ (ab + bc + ca) b2 c2 c2 a2 a2 b2 + + a (b + c) b (c + a) c (a + b) On the other hand, we have −→ −−→ OA · OB = ab + bc + ca Thus c2 a2 a2 b2 3√ b2 c2 + + ≥ (ab + bc + ca) ≥ a2 b2 c2 = , a (b + c) b (c + a) c (a + b) 2 which leads us to the inequality (1) Generalizations of the inequality (1) Theorem Let a, b, c be positive real numbers such that abc = 1, and let λ ≥ Then 1 + + ≥ aλ (b + c) bλ (c + a) cλ (a + b) Proof Let a = x1 , b = y1 , c = z1 Then 1 xλ−1 y λ−1 z λ−1 + λ + λ = + + , + c) b (c + a) c (a + b) y+z z+x x+y aλ (b By symmetry, we may assume that x ≥ y ≥ z, then xλ−2 ≥ y λ−2 ≥ z λ−2 , and 1 ≥ ≥ y+z z+x x+y xλ−2 y λ−2 z λ−2 ≥ ≥ y+z z+x x+y Using the rearrangement inequality gives xλ−1 y λ−1 z λ−1 xλ−2 y λ−2 z λ−2 + + ≥z· +x· +y· , y+z z+x x+y y+z z+x x+y xλ−1 y λ−1 z λ−1 xλ−2 y λ−2 z λ−2 + + ≥y· +z· +x· y+z z+x x+y y+z z+x x+y Adding the above inequalities yields xλ−1 y λ−1 z λ−1 + + y+z z+x x+y ≥ ≥ = λ−2 (x + y λ−2 + z λ−2 ) 3 λ−2 λ−2 λ−2 x y z (2) The inequality (2) is proved Theorem Let x1 , x2 , , xn be positive real numbers such that x1 x2 · · · xn = 1, and let n ≥ 3, λ ≥ Then xi )λ−1 ( ( 1≤k 0, i = 1, 2, , n, p ≥ or p ≤ 0), we deduce that 1≤k