Proving the inequality

Question

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?

Answer 1

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

Answer 2

I have to rush off now; if I remember, I'll come back and finish this answer.

Combinatorially: I'll interpret everything as areas.

Say we have $n$ sticks of length $a$, and $m$ sticks of length $b$. Then $a^2$ is the area of the square we get by placing two $a$-length sticks orthogonally; there are $\binom{n}{2}$ ways to do this, so $\binom{n}{2} a^2$ is the sum of the areas of all the squares we can make this way.

Similarly, $\binom{m}{2} b^2$ is the sum of the areas of all the $b$-squares we can make.

On the other hand, the right-hand side is $\binom{n+m}{2}$ the number of ways we can pick two sticks, multiplied by the square of $\frac{na+mb}{n+m}$ the average length of a stick.

The left-hand side has counted all the area we can possibly make by picking two of the same kind of stick; while the right-hand side has counted every possible rectangle but has counted each as being of area "the square of the average side length" instead of each rectangle's true area.

Probability beckons! Divide through by $\binom{n+m}{2}$, so that the right-hand side becomes the square of the average length of a stick, while the left-hand side becomes "the total area we can make by picking two sticks of the same size, divided by the number of ways to pick two sticks of any size". That's a quantity less than or equal to "the total area we can make by picking two sticks of the same size, divided by the number of ways to pick two sticks of the same size", which is just the mean area of a rectangle made of two sticks of the same size.

So it would be enough to show that the mean area of a rectangle made of two sticks of the same size is less than or equal to the square of the mean length of a stick.