Equilateral triange, sum... Just a short question:
In a triangle we have $\sum \left(\frac{a}{b+c}\right)^{2}$. Is the triangle equilateral?
I have derived 

For all real numbers $x,y,z$ we have that
$3(x^2+y^2+z^2)\geq (x+y+z)^2$. (It follows from expanding $\sum (x-y)^2 \geq 0$)
Applying this to your sum, we have that
$\sum \left(\frac{a}{b+c}\right)^2 \geq \frac{1}{3}\left(\sum\frac{a}{b+c}\right)^2$ with equality iff $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$.
Now Nesbitt's Inequality gives that $\sum\frac{a}{b+c}\geq\frac{3}{2}$, and so we have that
$\sum \left(\frac{a}{b+c}\right)^2\geq\frac{1}{3}\left(\frac{3}{2}\right)^2=\frac{3}{4}$.
We are given that we have equality, so we must have equality in each of the inequalities used. In particular, the equality condition for Nesbitt's Inequality is $a=b=c$ and hence the triangle is equilateral.
