<s>[INST] How prove $((x-y)(y-z)(z-x))^2\le 2((x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2)$ let $x,y,z>0$, and such $$x^2+y^2+z^2=x^2y^2+y^2z^2+x^2z^2$$
show that
$$((x-y)(y-z)(z-x))^2\le 2((x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2)$$
My try: let
$$x-y=a,y-z=b,z-x=c\Longrightarrow a+b+c=0$$
then we only prove
$$2[a(a+b)^2+b(b+c)^2+c(c+a)^2]\ge a^2b^2c^2$$
or
$$2(ac^2+ba^2+cb^2)\ge a^2b^2c^2$$
then I can't
 [/INST] Indeed, we can prove the following sharper inequality holds
$$2\big[(x-y)(y-z)(z-x)\big]^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}$$
Without loss of generality, assume that $x \geq y \geq z > 0,$ then
\begin{aligned} 2\big[(x-y)(y-z)(z-x)\big]^{2} \ & \leq 2(x-y)^2x^2y^2 \leq 2(x-y)^2(x^2y^2+y^2z^2+z^2x^2) \\ & =2(x-y)^2(x^2+y^2+z^2) \leq 2(x-y)^2(x^2+y^2+2xy) \\ & =2(x^2-y^2)^2 \leq (x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2.\end{aligned}
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