<s>[INST] 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.
 [/INST] 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.
</s>