<s>[INST] Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$ Let $a,b,c>0$ prove or disprove
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
since
$$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+a^2c^2)\tag{1}$$
other 
$$(a+b+c)^2\le 3(a^2+b^2+c^2)\tag{2}$$
Because of the inequality sign in a different direction so we can't $(1)\times (2)$   
 [/INST] After a quick look:
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge \sqrt[3]{\dfrac{(2s+c)(2s^2+c^2)}{9}} \ge \sqrt[4]{\dfrac{(s^4+2s^2c^2)}{3}} \ge \sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
where $s=\frac{a+b}{2}$ and we assume $c=\min(a,b,c)$.
The left inequality follows from $2(a^2+b^2)\ge (a+b)^2$.
The right inequality follows from $(s^2-p)(s^2+p-2c^2)\ge 0$ where $p=ab$ (note that $s^2\ge p \ge c^2$).
The middle inequality is, as you may have observed, the original inequality where we set $a=b=s$. Since the inequality is homogeneous, we can set $s=1$, i.e. solving the original inequality when $a=b=1$, which is easy.
This method is called Mixing Variables Method.
You can also try to find an intermediate term like in this solution to get an even nicer solution.
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