crazyfehmy wrote: Show that for all real numbers $x, y$ satisfying $x+y \geq 0$ \[ (x^2+y^2)^3 \geq 32(x^3+y^3)(xy-x-y) \] We can solve the problem by deriving the function $f(x,y) = RHS - LHS$. Details are given below. Let $v = \sqrt{x^2 + y^2}$. Then \[ x^3 + y^3 \le 2(x^2+y^2)^{3/2} = 2v^3 \] And \[ xy-x-y \le xy \le \frac{1}{2}(x^2+y^2) = \frac{v^2}{2} \] Hence $RHS \le 32v^5$. So if $v \ge 32$ we are done. Let $A$ be given by: \[ A = \{ (x,y) \in \mathbb{R}^2 \mid x+y \ge 0, x^2+y^2 \le 32^2 \} \] And then define $f$: \[ f(x, y) = (x^2+y^2)^3 - 32(x^3+y^3)(xy-x-y) \] We need only to show that $f(x,y) \ge 0$. If $x+y = 0$ then $f(x,y) = (x^2+y^2)^3 \ge 0$, and if $x^2+y^2 = 32^2$ then $f(x,y) \ge 0$ 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: \[ \frac{\partial f}{\partial x}(x,y) = \frac{\partial f}{\partial y}(x,y) = 0 \] Let $(x,y)$ be such a point. \[ \frac{\partial f}{\partial x}(x,y) = 6x(x^2+y^2)^2 - 96x^2(xy-x-y) - 32(x^3+y^3)(y-1) = 0 \] \[ \frac{\partial f}{\partial y}(x,y) = 6y(x^2+y^2)^2 - 96y^2(xy-x-y) - 32(x^3+y^3)(x-1) = 0 \] So: \[ (6x-6y)(x^2+y^2)^2 - (96x^2-96y^2)(xy-x-y) - 32(x^3+y^3)(y-x) = 0 \] So assume that $x \neq y$. Then: \[ 3(x^2+y^2)^2 - 48(x+y)(xy-x-y) + 16(x^3+y^3) = 0 \] Or equivalently \[ 3(x^2+y^2)^2 + 48(x+y)^2 + 16(x^3+y^3) = 48(x+y)xy \] Let $a = \frac{x+y}{2}$. Then \[ LHS \ge 12a^4 + 192a^2 + 32a^3, RHS \le 96a^3 \] Since $a > 0$ this is equivalent to $3a^2 - 16a + 48 \le 0$. Rewriting gives: \[ (3a - 8)^2 + 80 \le 0 \] This is clearly impossible. Hence $x = y$. But we can easily check the original inequality with $x = y$. Then it is equivalent to: \[ 8x^6 \ge 64x^3(x^2-2x) \] Or: \[ x^6 - 8x^5 + 16x^4 \ge 0 \] But the left hand side is $x^4(x-4)^2$, which is clearly non-negative. So the problem is solved.

crazyfehmy wrote: Show that for all real numbers $x, y$ satisfying $x+y \geq 0$ \[ (x^2+y^2)^3 \geq 32(x^3+y^3)(xy-x-y) \] $(x^2+y^2)^3 \geq 32(x^3+y^3)(xy-x-y)\Leftrightarrow$ $\Leftrightarrow(x^2+y^2)^3 + 32(x^3+y^3)(x+y)\geq32xy(x^3+y^3)$. But $(x^2+y^2)^3 + 32(x^3+y^3)(x+y)\geq2\sqrt{32(x^2+y^2)^3 (x^3+y^3)(x+y)}$. Hence, it remains to prove that $2\sqrt{32(x^2+y^2)^3 (x^3+y^3)(x+y)}\geq32xy(x^3+y^3)$, which is easy.

crazyfehmy wrote: Show that for all real numbers $x, y$ satisfying $x+y \geq 0$ \[ (x^2+y^2)^3 \geq 32(x^3+y^3)(xy-x-y) \] If $xy-x-y<0$, then inequality is ovious because $LHS \ge 0 \ge RHS$, so we just need to consider case when $xy-x-y \ge 0$. Using Am-Gm we have: $8 \cdot 2(x+y)\cdot (x^2-xy+y^2) \cdot 2(xy-x-y) \leq \frac{8}{27} (x^2+xy+y^2)^3$ Hence, it remains to prove: $\frac{8}{27} (x^2+xy+y^2)^3 \leq (x^2+y^2)^3$ What follows from: $2(x^2+xy+y^2) \leq 3(x^2+y^2) \Leftrightarrow (x-y)^2 \ge 0$. Equality holds when $x=y=4$. $\blacksquare$

let $x+y=p>0$ and $xy=q$ then if $q\leq p$ the statement will be obvious then assume that $q-p>0$ the statement will become as follows : $(p^{2}-2q)^{3}\geq 32(p)(p^{2}-3q)(q-p)$ then we must to prove that $3(p^{2}-2q)\geq 3\sqrt[3]{4p.2(p^{2}-3q).4(q-p)}$ by $AM-GM$. RHS is less than or equal to $4p+2p^{2}-6q+4q-4p$ then we must to prove that $3p^{2}-6q\geq4p+2p^{2}-6q+4q-4p$ which is equivalent to $p^{2}\geq 4q$ the final statement is obvious so we are done

If $xy\leq x+y,$ then $LHS\geq 0\geq (x+y)(x^2-xy+y^2)(xy-x-y)$ so we can suppose $xy\geq x+y\geq 0\implies x,y\geq 0$. After saying $x=4a,y=4b$ \[(16a^2+16b^2)^3\overset{?}{\geq}32(64a^3+64b^3)(16ab-4a-4b)\]\[\frac{LHS-RHS}{2^{12}}= a^6+3a^4b^2+3a^2b^4+b^6+2a^4+2b^4+2a^3b+2ab^3-8a^4b-8ab^4\]\[=(a^4+b^4)(a-b)^2+2a^4(b-1)^2+2b^4(a-1)^2+2a^3b(a-1)^2+2ab^3(b-1)^2\geq 0 \]

$f(x)=x^6 + 3 x^4 y^2 + 3 x^2 y^4 + y^6-32 x^4 y + 32 x^4 + 32 x^3 y - 32 x y^4 + 32 x y^3 + 32 y^4$ $f'(x)=6x^5+12x^3y^2+6xy^3-128x^3y+128x^3+96x^2y-32y^4+32y^3=0$ $g'(y)=6y^5+12y^3x^2+6y^3x-128y^3x+128y^3+96y^2x-32x^4+32x^3$ $-g'(y)=-6y^5-12y^3x^2-6y^3x+128y^3x-128y^3-96y^2x+32x^4-32x^3$ $0=f'-g'=6(x^5-y^5)+12x^2y^2(y-x)+128xy(y^2-x^2)+128(x^3-y^3)+96xy(x-y)+32(x^4-y^4)+32(y^3-x^3)=0$ $0=(x-y)(6A-128xy+128(x^2+xy+y^2)+96xy+32((x+y)(x^2+y^2))-32(x^2+xy+y^2)=0$ $(x-y)(x^4y+x^3y^2+x^2y^3+xy^4+128x^2+128y^2+32x^3+32xy^2+32yx^2+32y^3-32x^2-32xy-32y^2)=0$ $x^4y+x^3y^2+y^3x^2+xy^4+96x^2+96y^2+ 32x^3+32xy^2+32yx^2+32y^3-32xy\ge 0$