<s>[INST] Prove $\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$ If $a,b,c>0$, prove that:
$$\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
My try: I used $a^4+b^4 \geq ab(a^2+b^2)$ to get:
$$(a^4+b^4)(a^4+c^4)(b^4+c^4) \geq a^2b^2c^2(a^2+b^2)(a^2+c^2)(b^2+c^2)$$
I am stuck with this inequality:
$$\sqrt[3]{(a^2+b^2)(a^2+c^2)(b^2+c^2)}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
I am not sure if it is true or not.
 [/INST] It's wrong.
Let $a=b=c.$
Thus, we have $$2a^2\geq\sqrt{\frac{(a^3+a^2)^3}{1+a^3}}$$ or
$$2\geq a\sqrt{\frac{(a+1)^2}{a^2-a+1}},$$ which is wrong for $a\rightarrow+\infty$.
</s>