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.