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 $a=b=c$.

Question

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.

Answer 1

$\displaystyle\left[\frac{2(ab+bc+ca)}{a+b+c}\right]^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b-----(1)$ 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

$\displaystyle{\frac{2(a+b+c)}{3}}^{a+b+c}>{\frac{2(ab+bc+ca)}{a+b+c}}^{a+b+c}>(a+b)^c\cdot (b+c)^a\cdot (c+a)^b$

Answer 2

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 $$\sum_{cyc}(a-b)^2\geq0.$$