Algebraic inequality with logarithms

Question

Prove $\forall a,b,c\in \langle1,+\infty\rangle$ /corrected/ $$2\Bigg(\frac{\log_b{a}}{a+b}+\frac{\log_c{b}}{b+c}+\frac{\log_a{c}}{c+a}\Bigg)\geq\frac{9}{a+b+c}$$ I think I have to rearrange the expression to get an $A-H$ inequality using the reciprocal values of both sides so as to change the $\geq\;$ into $\;\leq$.

I also considered something like this groupping: $$\frac{\Bigg(\frac{\log_b{a}}{a+b}+\frac{\log_c{b}}{b+c}\Bigg)+\Bigg(\frac{\log_a{c}}{c+a}+\frac{\log_b{a}}{a+b}\Bigg)+\Bigg(\frac{\log_c{b}}{b+c}+\frac{\log_a{c}}{c+a}\Bigg)}{3}\geq\frac{3}{a+b+c}$$ $$\frac{a+b+c}{3}\geq\frac{3}{\Bigg(\frac{\log_b{a}}{a+b}+\frac{\log_c{b}}{b+c}\Bigg)+\Bigg(\frac{\log_a{c}}{c+a}+\frac{\log_b{a}}{a+b}\Bigg)+\Bigg(\frac{\log_c{b}}{b+c}+\frac{\log_a{c}}{c+a}\Bigg)}$$

I also tried the base changing: $$\log_b{a}=\frac{\log_c{a}}{\log_c{b}},\;\;\log_c{a}=\frac{1}{\log_a{c}}$$ $$\implies\log_b{a}=\frac{1}{\log_a{c}\times\log_c{b}}\implies\log_b{a}\times\log_a{c}\times\log_c{b}=1$$

This is all I have done so far.

Task from the republic contest in Yugoslavia, 1976

Answer 1

It's wrong. Try $a=0$.

For $a>1$, $b>1$ and $c>1$ by AM-GM we obtain: $$\sum_{cyc}(a+b)\sum_{cyc}\frac{\ln{a}}{(a+b)\ln{b}}\geq9\sqrt[3]{\prod_{cyc}(a+b)\prod_{cyc}\frac{\ln{a}}{(a+b)\ln{b}}}=9.$$

Answer 2

I think you are missing some constraint.

1. How do you define the $\log$ function when $a,b,c<0$?
2. Even if with the assumption $a,b,c>0$, I can get the following counter example:

$a=10^{-4}, b=10, c=10$

$$ \frac{\log_b(a)}{a+b} \approx \frac{-4}{10} = -0.4 $$ $$ \frac{\log_c(b)}{b+c} = \frac{1}{20} = 0.05 $$ $$ \frac{\log_a(c)}{a+c} \approx \frac{-0.25}{10} = -0.025 $$

so $$ \text{LHS} \approx 2(-0.4+0.05-0.025) < 0 < \frac{9}{20.0001} = \text{RHS} $$

Thus, I guess the constraint should be at least $a,b,c>1$