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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.

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    Do you know which contest is this problem from? Does the inequality have any particular motivation? – 2017-01-27
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    My personal curiosity is for the inspiration for the inequality, what it might be related to, what other areas of mathematics it might be useful in (perhaps some field of analysis?). That sort of information would transform the question from just a problem into a deeper and more broadly interesting post. I do see the link to AOPS above, but there is also no information there about the source or inspiration for the inequality, and nothing about what else it could be related to. – 2017-01-28
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    I am sure that the posts there are inspired by contest problems, but I think that before a question is ready to post here, we need to know more about it than just the fact that it happened to be posted at AOPS. As I was saying, my own curiosity is for what other inequalities this one is related to, how it could be applied, etc. - as it stands on AOPS is has no motivation, and so simply copying such inequalities here can lead to posts that lack motivation. Math.SE is not intended, in my opinion, as a forum to merely post contest-like problems for others to solve; AOPS may be more suitable. – 2017-01-28

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