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 } $

$$ f(x) = \frac{1}{1-x}-\frac{2}{1+x} $$ 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.

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.