<s>[INST] 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.
 [/INST] $\displaystyle\left[\frac{2(ab+bc+ca)}{a+b+c}\right]^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b-----(1)$
unless $a=b=c$
You can continue from here as follows
Note that $a^2+b^2+c^2> ab+bc+ca$
It follows that $\displaystyle\frac{(a+b+c)^2}{3}>ab+bc+ca$ and equivalently
$\implies\displaystyle\frac{2(a+b+c)^2}{3}>2(ab+bc+ca)$
$\implies\displaystyle\frac{2(a+b+c)}{3}>\frac{2(ab+bc+ca)}{a+b+c}$
Thus you have proved
$\displaystyle{\frac{2(a+b+c)}{3}}^{a+b+c}>{\frac{2(ab+bc+ca)}{a+b+c}}^{a+b+c}>(a+b)^c\cdot (b+c)^a\cdot (c+a)^b$
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