Check that \[ x = \frac{\ln b + \ln c}{\ln a} = \frac{-\ln b -\ln c}{-\ln a}. \]After setting $u=-\ln a$, $v=-\ln b$ and $w=-\ln c$, it boils down proving \[ \sum \frac{2w}{u+v+2w}\le \frac32 \iff \sum\left(1-\frac{u+v}{u+v+2w}\right)\le \frac32, \]which is equivalent to proving \[ \sum \frac{u+v}{u+v+2w}\ge \frac32. \]Finally, by Cauchy-Schwarz, \[ \sum \frac{u}{u+v+2w}\ge \frac{(u+v+w)^2}{\sum u^2 + 3(uv+vw+wu)} \quad\text{and}\quad \sum \frac{v}{u+v+2w}\ge \frac{(u+v+w)^2}{\sum u^2 + 3(uv+vw+wu)}. \]Combining, and realizing \[ \frac{2(u+v+w)^2}{\sum u^2 + 3(uv+vw+wu)} \ge \frac{3}{2} \]we are done.

grupyorum wrote: Check that \[ x = \frac{\ln b + \ln c}{\ln a} = \frac{-\ln b -\ln c}{-\ln a}. \]After setting $u=-\ln a$, $v=-\ln b$ and $w=-\ln c$, it boils down proving \[ \sum \frac{2w}{u+v+2w}\le \frac32 \iff \sum\left(1-\frac{u+v}{u+v+2w}\right)\le \frac32, \]which is equivalent to proving \[ \sum \frac{u+v}{u+v+2w}\ge \frac32. \]Finally, by Cauchy-Schwarz, \[ \sum \frac{u}{u+v+2w}\ge \frac{(u+v+w)^2}{\sum u^2 + 3(uv+vw+wu)} \quad\text{and}\quad \sum \frac{v}{u+v+2w}\ge \frac{(u+v+w)^2}{\sum u^2 + 3(uv+vw+wu)}. \]Combining, and realizing \[ \frac{2(u+v+w)^2}{\sum u^2 + 3(uv+vw+wu)} \ge \frac{3}{2} \]we are done. What's the point at multiplying the first equation by -1?

Tellocan wrote: What's the point at multiplying the first equation by -1? $0<a<1 \to \ln{a}<0$ so it was replaced with positive number

grupyorum wrote: \[\sum \frac{u+v}{u+v+2w}\ge \frac32 \] https://artofproblemsolving.com/community/c6h2588521p26753044: $$\frac{2\sqrt{ab}}{a+b+2c}+\frac{2\sqrt{bc}}{b+c+2a}+\frac{2\sqrt{ca}}{c+a+2b} \le \frac {3}{2} \le \frac{a+b}{a+b+2c} + \frac{b+c}{b+c+2a} + \frac{c+a}{c+a+2b}$$

RagvaloD wrote: Tellocan wrote: What's the point at multiplying the first equation by -1? $0<a<1 \to \ln{a}<0$ so it was replaced with positive number Ah i see thanks