<s>[INST] $\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$ 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.
 [/INST] 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)} $
</s>