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 $(1)\times (2)$   

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\sqrt[3]{u(3u^2-2v^2)}\geq\sqrt[4]{3v^4-2uw^3}$$ or
$f(w^3)\geq0$, where $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $w^3$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or
$x^3-3ux^2+3v^2x=w^3$. 
Hence, a line $y=w^3$ and a graph of $y=x^3-3ux^2+3v^2x$ have three common points.
Thus, $w^3$ gets a minimal value, when a line $y=w^3$ is a tangent line 
to the graph of $y=x^3-3ux^2+3v^2x$, which happens for equality case of two variables.
Also we must check the case $w^3\rightarrow0^+$.
Id est, it's enough to prove our inequality in the following cases.


*

*$b=c=1$, which gives 


$$(a-1)^2(a^{10}+10a^9+51a^8+188a^7+557a^6+1374a^5+983a^4+1616a^3+501a^2+538a+13)\geq0$$
2. $w^3\rightarrow0^+$. 
Let $c\rightarrow0^+$ and $b=1$. 
We obtain $(a^3+a^2+a+1)^4\geq243a^6$, which is obvious by AM-GM.
Done!
