show that $\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $ Let $a,b,c$  are positive numbers,if $$a+b+c=1$$

show that $$\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $$

I am tried proving it but failed.Any hints will be appreciated.
 A: $$ f(x) = \frac{1}{1-x}-\frac{2}{1+x} $$
is not a convex function on $(0,1)$, since:
$$ f''(x) = \frac{2}{(1-x)^3}-\frac{4}{(1+x)^3} \geq 0 $$
is equivalent to:
$$ \frac{1+x}{1-x}\geq \sqrt[3]{2} $$
but if we consider that $f'\left(\frac{1}{3}\right)=\frac{27}{8}$, it is not difficult to prove the algebraic inequality
$$ \forall x\in[0,1],\qquad f(x) \geq \frac{27}{8}\left(x-\frac{1}{3}\right) \tag{1}$$
since $f(x)-\frac{27}{8}\left(x-\frac{1}{3}\right)=0$ is equivalent to $\left(x-\frac{1}{3}\right)^2\left(x+\frac{1}{3}\right)=0$. 
Given $(1)$ and $a,b,c\in[0,1]$, $\,a+b+c=1$, it follows that:

$$ f(a)+f(b)+f(c) \geq \frac{27}{8}(a+b+c-1) = 0\tag{2} $$

as wanted.
A: We shall use the sigma sign ($\sum$) for cyclic sum.
By changing every number 1 into $a+b+c$, what we are trying to prove is equivalent to :
$$\sum \frac {1}{a+b} \ge \sum \frac{2}{(a+b)+(c+a)}$$
Let 
$$a+b=x$$
$$b+c=y$$
$$c+a=z$$
We have to prove :
$$\sum \frac {1}{x} \ge \sum \frac{2}{x+y}$$
We have $$ \frac 1x + \frac 1y \ge \frac 4 {x+y}$$
$$ \frac 1y + \frac 1z \ge \frac 4 {y+z}$$
$$ \frac 1z + \frac 1x \ge \frac 4 {z+x}$$
 By adding all three side of all three inequalities above, we get 
$$\sum \frac {2}{x} \ge \sum \frac{4}{x+y}$$
Dividing each side by 2, we get what we have to prove.
