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.

  • 1
    $\begingroup$ Do you know which contest is this problem from? Does the inequality have any particular motivation? $\endgroup$ Jan 27, 2017 at 2:00
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 28, 2017 at 14:42
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 28, 2017 at 15:18

1 Answer 1


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.

  • $\begingroup$ I went through the link but I find difficult to follow the lemma. When $0<p<1$ what do we need to show that the function $F$ exists and is differentiable ? On the same foot, how can we write $a_i$ (not a symmetric polynomial) as a function of the symmetric polynomials, as it seems it is done when differentiating with respect to $E_r$ ? $\endgroup$
    – Thomas
    Jul 3, 2019 at 18:09
  • $\begingroup$ @Thomas We work with $f(t)=t^p,$ where $0<p<1$. In out case $p=\frac{1}{4}.$ $\endgroup$ Jul 3, 2019 at 18:29
  • $\begingroup$ Thanks! But my questions were more about the proof in the link you wrote rather than in the application of the result. Or maybe I did no understand your answer... $\endgroup$
    – Thomas
    Jul 3, 2019 at 18:52
  • $\begingroup$ @Thomas You can see a proof in the link. Ask this question in AoPS. Also, you can open a new topic with this question in this forum. $\endgroup$ Jul 3, 2019 at 19:17
  • 1
    $\begingroup$ @Thomas In my opinion, some details are missing in Ji Chen's proof (for example, the derivatives do not exist if some variables are equal). But it is not hard to come up with a rigorous proof for Ji Chen's lemma for $n=3$ case. I think that it is interesting to discuss the rigorous proof of Ji Chen's lemma. $\endgroup$
    – River Li
    Sep 7, 2019 at 7:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .