$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$ I tried to solve this inequality:
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$$
with 
$$a+b+c+d+e=1$$
I am stuck at this. I don't want the full solution, a hint would be enough.
 A: Hint:
You have $1-a=b+c+d+e$ and
$a=1-(b+c+d+e)$
$\dfrac{1-(b+c+d+e)}{b+c+d+e}=\dfrac{1}{b+c+d+e}-1$ and $b+c+d+e=1-a$
Using AM $\ge$ HM
$(\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}+\dfrac{1}{1-d}+\dfrac{1}{1-e}) \ge \dfrac{5^2}{1+1+1+1+1-(a+b+c+d+e)} $
A: HINT:
$$\frac a{1-a}=-1+\frac1{1-a}$$
Assuming  $a,b,c,d,e$ to be positive real numbers, $1-a>0 $
Using AM HM inequality on $1-a$ etc 
$$\frac{\sum (1-a)}5\ge \frac5{\sum \frac1{1-a}}\implies \sum \frac1{1-a}\ge\frac{5^2}{5-\sum a}=\frac{25}{5-1}$$
A: Hint: with $g(a,b,c,d,e) = a + b + c + d + e$ and 
$$f(a,b,c,d,e) = \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}$$
we can consider optimizing $f$ over the set of $(a,b,c,d,e)$ for which $g(a,b,c,d,e) = 1$. The method of lagrange multipliers gives a necessary condition for the existence of a critical point on this set: $\nabla f = \lambda \nabla g$ for some $\lambda \in \mathbb{R}$.
However we know that $\nabla g$ is the column vector $(1,1,1,1,1)$, which means that
$$
\frac{1}{(1-x)^2} = \lambda
$$
for each $x = a,b,c,d,e$ individually. Notice that this condition is the same for each coordinate. 
What can you conclude about the solutions to the lagrange multiplier problem, and how can you connect this information to the inequality you wish to prove?
A: A proof in one line:
$$\sum_{cyc}\left(\frac{a}{1-a}-\frac{1}{4}\right)=\sum_{cyc}\left(\frac{(5a-1)}{4(1-a)}-\frac{5}{16}(5a-1)\right)=\sum_{cyc}\frac{(5a-1)^2}{16(1-a)}\geq0.$$
Done!
