Prove that $\sum\limits_{cyc}\sqrt[4]{a^2+bc+ca}\geq3\sqrt[4]{ab+ac+bc}$ Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\sqrt[4]{a^2+bc+ca}+\sqrt[4]{b^2+ca+ab}+\sqrt[4]{c^2+ab+bc}\geq3\sqrt[4]{ab+ac+bc}$$
I tried Holder and more, but without any success. 
I didn't think that my delirium would interesting to someone.
OK. By Holder 
$$\left(\sum\limits_{cyc}\sqrt[4]{a^2+bc+ca}\right)^4\sum\limits_{cyc}(a^2+bc+ca)^4(ka+mb+c)^5\geq$$
$$\geq\left(\sum\limits_{cyc}(a^2+bc+ca)(ka+mb+c)\right)^5$$
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a^2+bc+ca)(ka+mb+c)\right)^5\geq81\sum\limits_{cyc}(a^2+bc+ca)^4(ka+mb+c)^5(ab+ac+bc)$$
and I did not find a non-negative values of $k$ and $m$, for which this inequality would true.

Now, I have a solution.
By using of the Ji Chen's lemma ( https://artofproblemsolving.com/community/c6h194103 )
it's enough to prove that
$$\sum_{cyc}(a^2+bc+ca)\geq3(ab+ac+bc),$$
$$\sum_{cyc}(a^2+bc+ca)(b^2+ca+ab)\geq3(ab+ac+bc)^2$$ and
$$\prod_{cyc}(a^2+bc+ca)\geq(ab+ac+bc)^3,$$ which are obvious.
