Show that for all real numbers x,y satisfying x+y≥0 (x2+y2)3≥32(x3+y3)(xy−x−y)
Problem
Source: Turkey JBMO Team Selection Test Problem 7
Tags: inequalities, function, inequalities proposed
30.05.2012 12:00
crazyfehmy wrote: Show that for all real numbers x,y satisfying x+y≥0 (x2+y2)3≥32(x3+y3)(xy−x−y) We can solve the problem by deriving the function f(x,y)=RHS−LHS. Details are given below. Let v=√x2+y2. Then x3+y3≤2(x2+y2)3/2=2v3 And xy−x−y≤xy≤12(x2+y2)=v22 Hence RHS≤32v5. So if v≥32 we are done. Let A be given by: A={(x,y)∈R2∣x+y≥0,x2+y2≤322} And then define f: f(x,y)=(x2+y2)3−32(x3+y3)(xy−x−y) We need only to show that f(x,y)≥0. If x+y=0 then f(x,y)=(x2+y2)3≥0, and if x2+y2=322 then f(x,y)≥0 as well. Since A is compact and f is continuous, f has a minimum. Since f is differentiable it suffices to check points (x,y) such that: ∂f∂x(x,y)=∂f∂y(x,y)=0 Let (x,y) be such a point. ∂f∂x(x,y)=6x(x2+y2)2−96x2(xy−x−y)−32(x3+y3)(y−1)=0 ∂f∂y(x,y)=6y(x2+y2)2−96y2(xy−x−y)−32(x3+y3)(x−1)=0 So: (6x−6y)(x2+y2)2−(96x2−96y2)(xy−x−y)−32(x3+y3)(y−x)=0 So assume that x≠y. Then: 3(x2+y2)2−48(x+y)(xy−x−y)+16(x3+y3)=0 Or equivalently 3(x2+y2)2+48(x+y)2+16(x3+y3)=48(x+y)xy Let a=x+y2. Then LHS≥12a4+192a2+32a3,RHS≤96a3 Since a>0 this is equivalent to 3a2−16a+48≤0. Rewriting gives: (3a−8)2+80≤0 This is clearly impossible. Hence x=y. But we can easily check the original inequality with x=y. Then it is equivalent to: 8x6≥64x3(x2−2x) Or: x6−8x5+16x4≥0 But the left hand side is x4(x−4)2, which is clearly non-negative. So the problem is solved.
30.05.2012 17:29
crazyfehmy wrote: Show that for all real numbers x,y satisfying x+y≥0 (x2+y2)3≥32(x3+y3)(xy−x−y) (x2+y2)3≥32(x3+y3)(xy−x−y)⇔ ⇔(x2+y2)3+32(x3+y3)(x+y)≥32xy(x3+y3). But (x2+y2)3+32(x3+y3)(x+y)≥2√32(x2+y2)3(x3+y3)(x+y). Hence, it remains to prove that 2√32(x2+y2)3(x3+y3)(x+y)≥32xy(x3+y3), which is easy.
30.05.2012 22:29
crazyfehmy wrote: Show that for all real numbers x,y satisfying x+y≥0 (x2+y2)3≥32(x3+y3)(xy−x−y) If xy−x−y<0, then inequality is ovious because LHS≥0≥RHS, so we just need to consider case when xy−x−y≥0. Using Am-Gm we have: 8⋅2(x+y)⋅(x2−xy+y2)⋅2(xy−x−y)≤827(x2+xy+y2)3 Hence, it remains to prove: 827(x2+xy+y2)3≤(x2+y2)3 What follows from: 2(x2+xy+y2)≤3(x2+y2)⇔(x−y)2≥0. Equality holds when x=y=4. ◼
01.06.2012 21:01
let x+y=p>0 and xy=q then if q≤p the statement will be obvious then assume that q−p>0 the statement will become as follows : (p2−2q)3≥32(p)(p2−3q)(q−p) then we must to prove that 3(p2−2q)≥33√4p.2(p2−3q).4(q−p) by AM−GM. RHS is less than or equal to 4p+2p2−6q+4q−4p then we must to prove that 3p2−6q≥4p+2p2−6q+4q−4p which is equivalent to p2≥4q the final statement is obvious so we are done
17.04.2024 19:03
If xy≤x+y, then LHS≥0≥(x+y)(x2−xy+y2)(xy−x−y) so we can suppose xy≥x+y≥0⟹x,y≥0. After saying x=4a,y=4b (16a2+16b2)3?≥32(64a3+64b3)(16ab−4a−4b)LHS−RHS212=a6+3a4b2+3a2b4+b6+2a4+2b4+2a3b+2ab3−8a4b−8ab4=(a4+b4)(a−b)2+2a4(b−1)2+2b4(a−1)2+2a3b(a−1)2+2ab3(b−1)2≥0
29.12.2024 20:21
f(x)=x6+3x4y2+3x2y4+y6−32x4y+32x4+32x3y−32xy4+32xy3+32y4 f′(x)=6x5+12x3y2+6xy3−128x3y+128x3+96x2y−32y4+32y3=0 g′(y)=6y5+12y3x2+6y3x−128y3x+128y3+96y2x−32x4+32x3 −g′(y)=−6y5−12y3x2−6y3x+128y3x−128y3−96y2x+32x4−32x3 0=f′−g′=6(x5−y5)+12x2y2(y−x)+128xy(y2−x2)+128(x3−y3)+96xy(x−y)+32(x4−y4)+32(y3−x3)=0 0=(x−y)(6A−128xy+128(x2+xy+y2)+96xy+32((x+y)(x2+y2))−32(x2+xy+y2)=0 (x−y)(x4y+x3y2+x2y3+xy4+128x2+128y2+32x3+32xy2+32yx2+32y3−32x2−32xy−32y2)=0 x4y+x3y2+y3x2+xy4+96x2+96y2+32x3+32xy2+32yx2+32y3−32xy≥0