proving $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 2(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})$

Question

For $a+b+c=1; a,b,c>0$, prove that $$ \frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} \leq 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right). $$

Answer 1

since $$\dfrac{1+a}{1-a}=\dfrac{-(1-a)+2}{1-a}=-1+\dfrac{2}{1-a}=-1+\dfrac{2}{b+c}=-1+\dfrac{2a+2b+2c}{b+c}=1+\dfrac{2a}{b+c}$$ or $$\dfrac{1+a}{1-a}=\dfrac{a+b+c+a}{b+c}=\dfrac{b+c+2a}{b+c}=1+\dfrac{2a}{b+c}$$ so $$\Longleftrightarrow 2\sum_{cyc}\dfrac{b}{a}-\sum_{cyc}\dfrac{1+a}{1-a}=2\sum_{cyc}\dfrac{b}{a}-\sum_{cyc}\dfrac{2a}{b+c}-3=2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}-\dfrac{2a}{b+c}-\dfrac{2b}{a+c}+\dfrac{2c}{a+b}\right)-3$$

$$\Longleftrightarrow \sum_{cyc}\dfrac{ab}{c(a+b)}\ge\dfrac{3}{2}$$ By Cauchy-Schwarz inequality we have $$\sum_{cyc}\dfrac{ab}{c(a+b)}\sum_{cyc}abc(a+b)\ge (\sum_{cyc}ab)^2$$ $$\Longleftrightarrow \dfrac{(ab+bc+ac)^2}{abc(a+b)+abc(b+c)+abc(a+c)}\ge\dfrac{3}{2}$$ $$\Longleftrightarrow (ab+bc+ac)^2\ge 3abc$$

Becuase $$(ab+bc+ac)^2\ge 3abc(a+b+c)=3abc$$ This AM-GM inequality.By done!