Let $a,b,c>0$ s.t. $a+b+c=1$.
Show that $\frac {1+a}{1-a} + \frac{1+b}{1-b} +\frac{1+c}{1-c}\leq \frac{2a}{b}+\frac{2b}{c} +\frac{2c}{a}.$
My idea: I have to show that
3+ $\frac{2a}{1-a}+\frac{2b}{1-b}+\frac{2c}{1-c}\leq \frac{2a}{b}+\frac{2b}{c} +\frac{2c}{a}$ which is equivalent with $ \frac{3}{2}+ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\leq \frac{2a}{b}+\frac{2b}{c} +\frac{2c}{a}.$
I used Nesbitt' s inequality and it's enough to show that $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\leq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}.$
I notice that $\frac{4a}{b+c}\leq \frac{a}{b}+\frac{a}{c}$. Now I am stuck.
