Prove $\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$

Question

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.

Answer 1

Raising to the $6$th power, the inequality is equivalent with:

$$(1+abc)^3(a^4+b^4)^2(b^4+c^4)^2(c^4+a^4)^2 \geq a^4b^4c^4(a^3+bc)^3(b^3+ca)^3(c^3+ab)^3$$

This is wrong for large variables as Michael Rozenberg stated. However, I found that

$$(1+abc)^3(a^4+b^4)^2(b^4+c^4)^2(c^4+a^4)^2 \geq a^2b^2c^2(a^3+bc)^3(b^3+ca)^3(c^3+ab)^3$$

is true. Notice that, using Holder's inequality, we have:

$$(a^4+b^4)(a^4+c^4)(abc+1)\geq (a^3\sqrt[3]{bc}+bc\sqrt[3]{bc})^3=bc(a^3+bc)^3$$

and now multiply with the other two similar inequalities to complete it.

Answer 2

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 $a\rightarrow+\infty$.