Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$

Question

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 $(1)\times (2)$

Answer 1

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.

Answer 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.

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. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=1$.

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

Done!