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

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.
$\textbf{Proved.}$
