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Let $a,b$ and $c$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$

This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can't find this inequality have solve it,maybe it seem can use integral to solve it?

my attempts:

I took $p=3a+2b,$ $2a+2b+c$. $k=3$ and I wanted to use this integral: $$\dfrac{1}{p^k}=\dfrac{1}{\Gamma(k)}\int_{0}^{+\infty}t^{k-1}e^{-pt}dt$$ but I don't see how it helps.

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    $\begingroup$ I think it was a typo in the formulation. This inequality is true, but has a very ugly solution. The inequality with nice solution is the following. $\sum\limits_{cyc}\frac{a^3b}{(2a+3b)^3}\geq\sum\limits_{cyc}\frac{a^2bc}{(2a+2b+c)^3}.$ $\endgroup$ – Michael Rozenberg Apr 3 '19 at 4:59
  • $\begingroup$ can you post your nice solution with $\sum_{cyc}\dfrac{a^3b}{(2a+3b)^3}\ge\sum_{cyc}\dfrac{a^2bc}{(2a+2b+c)^3}$?Thanks $\endgroup$ – function sug Apr 3 '19 at 5:00
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    $\begingroup$ @MichaelRozenberg,I have add mu attempts,and take $p=3a+2b$,$2a+2b+c$.$k=3$,maybe can help? $\endgroup$ – function sug Apr 3 '19 at 5:41
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    $\begingroup$ This inequality is obviously true by BW, but this solution is not by hand. $\endgroup$ – Michael Rozenberg Apr 10 '19 at 15:10
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    $\begingroup$ Also, this inequality is obviously true after full expanding by AM-GM. But for the proof we need to use a computer again. $\endgroup$ – Michael Rozenberg Apr 10 '19 at 20:43
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The proof of my inequality.

let $a$, $b$ and $c$ be positive numbers. Prove that: $$\tfrac{a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+2c)^3}+\tfrac{c^3a}{(2c+3a)^3}\geq\tfrac{a^2bc}{(2a+2b+c)^3}+\tfrac{b^2ca}{(2b+2c+a)^3}+\tfrac{c^2ab}{(2c+2a+b)^3}.$$

Indeed, by Holder and AM-GM we obtain: $$\sum_{cyc}\tfrac{a^3b}{(2a+3b)^3}=\sum_{cyc}\tfrac{\left(4(2a+3b)+(2b+3c)+2(2c+3a)\right)^3\left(\tfrac{4a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+3c)^3}+\tfrac{2c^3a}{(2c+3a)^3}\right)}{2401(2a+2b+c)^3}\geq$$ $$\geq\sum_{cyc}\frac{\left(4\sqrt[4]{a^3b}+\sqrt[4]{b^3c}+2\sqrt[4]{c^3a}\right)^4}{2401(2a+2b+c)^3}\geq\sum_{cyc}\frac{\left(7\sqrt[28]{a^{12+2}b^{4+3}c^{1+6}}\right)^4}{2401(2a+2b+c)^3}=\sum_{cyc}\frac{a^2bc}{(2a+2b+c)^3}.$$

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Using binomial inequality for $$|c-a|<2a+2b+c,$$ one can get $$\dfrac1{(3a+2b)^3}=\dfrac1{(2a+2b+c)^3}\left(1-\dfrac{c-a}{2a+2b+c}\right)^{-3} \ge \dfrac1{(2a+2b+c)^3}\left(1+3\dfrac{c-a}{2a+2b+c}\right),$$ $$\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3} \ge \dfrac{a^2b}{(2a+2b+c)^3}\left(a-c-3a\dfrac{c-a}{2a+2b+c}\right),\tag1$$ The issue inequality can be presented in the form of $$S\ge 0\tag2,$$ where $$S = \sum\limits_\bigcirc \left(\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3}\right) \ge \sum\limits_\bigcirc\dfrac{a^2b(a-c)}{(2a+2b+c)^3} +3\sum\limits_\bigcirc\dfrac{a^3b(a-c)}{(2a+2b+c)^4}.$$

In according with the rearrangement inequality for the productions of $$a\cdot a\cdot a\cdot \dfrac b{2a+2b+c} + b\cdot b\cdot b\cdot \dfrac c{2b+2c+a} + c\cdot c\cdot c\cdot \dfrac a{2c+2a+b}$$ and $$a\cdot a\cdot c\cdot \dfrac b{2a+2b+c} + b\cdot b\cdot a\cdot \dfrac c{2b+2c+a} + c\cdot c\cdot b\cdot \dfrac a{2c+2a+b}$$ between $a$ and $c$ (where the ratio can be placed arbitrary), one can write $$\sum\limits_\bigcirc\dfrac{a^3b}{(2a+2b+c)^3} \ge \sum\limits_\bigcirc\dfrac{a^2bc}{(2a+2b+c)^3}$$ and similarly
$$\sum\limits_\bigcirc\dfrac{a^4b}{(2a+2b+c)^4} \ge \sum\limits_\bigcirc\dfrac{a^3bc}{(2a+2b+c)^4},$$
so $(1)$ is valid.

$\textbf{Proved.}$

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  • $\begingroup$ To which numbers exactly is the rearrangement inequality applied? $\endgroup$ – Martin R Apr 17 '19 at 12:03
  • $\begingroup$ @MartinR To subsequences $\{a,a\}$ and $\{c.c\}.$ $\endgroup$ – Yuri Negometyanov Apr 17 '19 at 12:08
  • $\begingroup$ Well, perhaps I am too dumb to see it. The rearrangement inequality states that $x_1 y_1 + x_2 y_2 + x_3 y_3$ is maximal if $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ are ordered in the same way (both ascending or both descending). What are $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ in your case so that the final two inequalities are obtained? $\endgroup$ – Martin R Apr 17 '19 at 12:22
  • $\begingroup$ @MartinR I think that the most clear way is to correspond the debominators with $b.$ $\endgroup$ – Yuri Negometyanov Apr 17 '19 at 12:31
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    $\begingroup$ @Yuri Negometyanov Can you write sequences, for which you used Rearrangement. Your sequences $(a,a)$ and $(c,c)$ give $a^2+c^2\geq2ac$, which is true, but it's nothing here. Actually, components of sequences should be ordered, otherwise Rearrangement does not work. $\endgroup$ – Michael Rozenberg Apr 17 '19 at 15:36

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