Vtyjkyuk

Let $a,b,c>0$ prove or disprove
$$sqrt[3]dfrac(a+b+c)(a^2+b^2+c^2)9gesqrt[4]dfrac(a^2b^2+b^2c^2+c^2a^2)3$$

since
$$(a^2+b^2+c^2)^2ge 3(a^2b^2+b^2c^2+a^2c^2)tag1$$

other
$$(a+b+c)^2le 3(a^2+b^2+c^2)tag2$$
Because of the inequality sign in a different direction so we can't $(1)times (2)$

After a quick look:
$$sqrt[3]dfrac(a+b+c)(a^2+b^2+c^2)9ge sqrt[3]dfrac(2s+c)(2s^2+c^2)9 ge sqrt[4]dfrac(s^4+2s^2c^2)3 ge sqrt[4]dfrac(a^2b^2+b^2c^2+c^2a^2)3$$
where $s=fraca+b2$ 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 $s^2ge p ge c^2$).

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.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that
$$sqrt[3]u(3u^2-2v^2)geqsqrt[4]3v^4-2uw^3$$ 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 $w^3rightarrow0^+$.

Id est, it's enough to prove our inequality in the following cases.

1. $b=c=1$, 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. $w^3rightarrow0^+$.

Let $crightarrow0^+$ and $b=1$.

We obtain $(a^3+a^2+a+1)^4geq243a^6$, which is obvious by AM-GM.

Done!