show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $

Question

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.

Answer 1

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: $$\sum_{cyc}\tfrac{a^3b}{(2a+3b)^3}=\sum_{cyc}\tfrac{\left(4(2a+3b)+(2b+3c)+2(2c+3a)\right)^3\left(\tfrac{4a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+3c)^3}+\tfrac{2c^3a}{(2c+3a)^3}\right)}{2401(2a+2b+c)^3}\geq$$ $$\geq\sum_{cyc}\frac{\left(4\sqrt[4]{a^3b}+\sqrt[4]{b^3c}+2\sqrt[4]{c^3a}\right)^4}{2401(2a+2b+c)^3}\geq\sum_{cyc}\frac{\left(7\sqrt[28]{a^{12+2}b^{4+3}c^{1+6}}\right)^4}{2401(2a+2b+c)^3}=\sum_{cyc}\frac{a^2bc}{(2a+2b+c)^3}.$$

Answer 2

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.

$\textbf{Proved.}$