<s>[INST] $a+b+c=1.$ 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.
 [/INST] Just another way, note for $x\in(0,1)$,
$$f(x)=\frac1{1-x}-\frac2{1+x}-\frac98(3x-1)=\frac{(3x-1)^2(3x+1)}{8(1-x^2)}\geqslant0$$
while the inequality is just $f(a)+f(b)+f(c)\geqslant 0$. 
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