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Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^4}{b}+\frac{b^4}{c}+\frac{c^4}{a}\geq a^3+b^3+c^3+25(a-b)(b-c)(c-a)$$

I tried the uvw's technique and BW and more but without some success.

For example, BW does not help here:

Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Hence, $$abc\left(\sum\limits_{cyc}\frac{a^4}{b}-\sum\limits_{cyc}a^3-25\prod\limits_{cyc}(a-b)\right)=4(u^2-uv+v^2)a^4+$$ $$+(6u^3+19u^2v-21uv^2+6v^3)a^3+(4u^4+21u^3v-19uv^3+4v^4)a^2+$$ $$+(u^5-u^4v+25u^3v^2-25u^2v^3+4uv^4+v^5)a+uv^5,$$ which is nothing.

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  • $\begingroup$ Interesting to note that $(a,b,c)$ satisfies the inequality if and only if $(\lambda a,\lambda b,\lambda c)$ satisfy the inequality for each $\lambda >0$. So it's fair game to assume that, say, $a=1$ and $0 <b,c \leq 1$. $\endgroup$ – Fimpellizieri Jan 2 '17 at 20:45
  • $\begingroup$ I tried using Polya's observation, that for a positive form $F$ and high enough $n$, $(a+b+c)^n F$ will have only positive coefficients. But even up to $n=60$ there are still some negatives. $\endgroup$ – Mark Fischler Jun 13 '17 at 18:21
  • $\begingroup$ @Mark Fischler The sixth degree cyclic homogeneous inequalities with three variables it's something very very hard. But symmetric sixth degree it's an easy enough. Thank you for your interest! $\endgroup$ – Michael Rozenberg Jun 13 '17 at 18:55
  • $\begingroup$ @Mark Fischler Polya's theorem may require very large $n$. $n=2000$ is not enough. $\endgroup$ – River Li Sep 7 at 15:41

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