$a+b+c=1.$ Show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$ 
Assume $a,b,c>0$ and $a+b+c=1.$ Show that
  $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq
 \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$

Here's what I tried:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+b}+\frac{a+b+c}{a+b} \geq
 \frac{1+a+b+c}{1+a}+\frac{1+a+b+c}{1+b}+\frac{1+a+b+c}{1+c}$
$\rightarrow \frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+b} \geq
 \frac{b+c}{1+a}+\frac{a+c}{1+b}+\frac{a+b}{1+c}$
Let $A=b+c, B=a+c,$ and $ C=a+b,$ so $A+B+C=2.$
$\rightarrow \frac{1-A}{A}+\frac{1-B}{B}+\frac{1-C}{C} \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
Or $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} - 3 \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
I am stuck here, thought about using AM-GM-HM to get rid of the reciprocal on the RHS but it doesn't work if applied directly.

From Jensen's Inequality with the convex function $f(t):= \frac{1}{1-t}$ for $t\in (0,1)$, we have
$$\frac{a}{1-c}f(a)+\frac{b}{1-c}f(b)\geq f\left(\frac{a^2}{1-c}+\frac{b^2}{1-c}\right)\,.$$
Since $f$ is increasing, the Power-Mean Inequality $\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2$ implies that
$$f\left(\frac{a^2+b^2}{1-c}\right)\geq f\left(\frac{(a+b)^2}{2(1-c)}\right)=f\left(\frac{1-c}{2}\right)\,.$$
Thus,
$$\frac{a}{1-a}+\frac{b}{1-b}\geq (1-c)\,f\left(\frac{a^2+b^2}{1-c}\right)\geq (1-c)\,f\left(\frac{1-c}{2}\right)=\frac{2(1-c)}{1+c}\,.$$
We have two more similar inequalities, and by summing all three of them (and dividing the result by $2$), we have
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1-a}{1+a}+\frac{1-b}{1+b}+\frac{1-c}{1+c}\,.$$
Adding $3$ to both sides of the inequality above, the required result is yielded.  The equality case is, of course, $(a,b,c)=\left(\frac13,\frac13,\frac13\right)$.
