Prove that $(a+b)^c\cdot(b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot (a+b+c)\right]^{a+b+c}$ where $a,b,c\in \mathbb{Q}^{+}$ unless $a=b=c$. 
If $a,b,c$ be positive rational numbers, prove that
$$(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot
(a+b+c)\right]^{a+b+c}$$ unless $a=b=c$

My try :
Consider $(a+b),(b+c),(c+a)$ be three numbers with associated weights $c,a,b$
By weighted AM-GM inequality,
$$\frac{c\cdot (a+b)+a\cdot (b+c)+b\cdot (c+a)}{a+b+c}\ge \biggl[(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\biggl]^\frac{1}{a+b+c}$$
the equality occurs when $(a+b)=(b+c)=(c+a)$ that is $a=b=c$
Therefore
$$\left[\frac{2(ab+bc+ca)}{a+b+c}\right]^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b-----(1)$$
unless $a=b=c$
Now $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
Therefore $$(a+b+c)^2\gt 2(ab+bc+ca)-------(2)$$
using equation $(1)$ and $(2)$, we get,
$$\left(a+b+c\right)^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b$$
I am stuck. please give me some hint. Thanks in advance.

We need to prove that:
$$\sum_{cyc}\frac{c}{a+b+c}\ln(a+b)\leq\ln\left(\frac{2}{3}(a+b+c)\right).$$
Now, $\sum\limits_{cyc}\frac{c}{a+b+c}=1$ and we can use Jensen for the concave function $\ln$.
Indeed, $$\sum_{cyc}\frac{c}{a+b+c}\ln(a+b)\leq\ln\left(\sum_{cyc}\frac{c(a+b)}{a+b+c}\right)$$ and it's enough to prove that
$$2(a+b+c)^2\geq3\sum_{cyc}c(a+b),$$ which is $$\sum_{cyc}(a-b)^2\geq0.$$
