Proving the inequality I need to prove this:
$\binom{n}{2}a^2 + \binom{m}{2}b^2 
\leq
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \right) ^2$
I tried this (WLOG $n\leq m$):
$\binom{n}{2}a^2 + \binom{m}{2}b^2 = \frac{n(n-1)}{2}a^2 + \frac{m(m-1)}{2}b^2
\\
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \right) ^2
= \frac{n+m-1}{2}\frac{(na+mb)^2}{n+m}$
How would you do it?

A full expanding gives $(a-b)^2+2(m+n)ab\geq0$ and the rest for you.
