Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}$

Question

Prove that $$\left( \frac{bc+ac+ab}{a+b+c} \right)^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$$ where $a,b,c>0$

My attempt:

I couldn't proceed after that. Please help me in this regard, thanks!

Answer 1

We need to prove that $$(a+b+c)\ln\frac{ab+ac+bc}{a+b+c}\geq\frac{1}{2}\sum\limits_{cyc}a\ln{bc}$$ or $$\ln\left(\frac{ab+ac+bc}{a+b+c}\right)^2\geq\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}.$$ But by Jensen $$\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}\leq\ln\sum\limits_{cyc}\frac{abc}{a+b+c}$$ Thus, it remains to prove that $$\frac{3abc}{a+b+c}\leq\left(\frac{ab+ac+bc}{a+b+c}\right)^2,$$ which is $$\sum\limits_{cyc}c^2(a-b)^2\geq0$$ Done!

Answer 2

$a^bb^cc^a\ge\sqrt{(ab)^c(bc)^a(ca)^b}$

$\implies a^{2b}b^{2c}c^{2a}\ge a^{b+c}b^{c+a}c^{a+b}$

$\implies a^bb^cc^a\ge a^cb^ac^b$

It doesn't seems true always because of simmatry..

Answer 3

If you want to use the weighted AM-GM inequality, considering

\begin{align*} \frac{bc+ca+ab}{a+b+c} &= \frac{2(bc+ca+ab)}{2(a+b+c)} \\ &= \frac{(b+c)\color{red}{\boldsymbol{a}}+ (c+a)\color{green}{\boldsymbol{b}}+ (a+b)\color{blue}{\boldsymbol{c}}}{(b+c)+(c+a)+(a+b)} \\ \left( \frac{bc+ca+ab}{a+b+c} \right)^{2(a+b+c)} &\ge \color{red}{\boldsymbol{a}}^{b+c} \color{green}{\boldsymbol{b}}^{c+a} \color{blue}{\boldsymbol{c}}^{a+b} \\ \left( \frac{bc+ca+ab}{a+b+c} \right)^{a+b+c} &\ge \sqrt{a^{b+c}b^{c+a}c^{a+b}} \\ &= \sqrt{(bc)^{a}(ca)^{b}(ab)^{c}} \end{align*}