# Prove an inequality

• Jun 6th 2007, 10:04 AM
puch7524
Prove an inequality
Can anyone prove the following inequality:

$\displaystyle \sum_{i=1}^N y_{i}^2/n_{i} - \frac{\left(\sum_{i=1}^N y_{i}\right)^2}{n} \ge 0,$

where $\displaystyle n=\sum_{i=1}^N n_{i},\ \ n_{i} \ge 0$, and $\displaystyle y_{i}$ is a real quantity (can be both negative or positive)?

It appears that the inequality is valid (even used random numbers), but can't see how to prove it.
• Jun 6th 2007, 11:56 AM
ThePerfectHacker
Quote:

Originally Posted by puch7524
Can anyone prove the following inequality:

$\displaystyle \sum_{i=1}^N y_{i}^2/n_{i} - \frac{\left(\sum_{i=1}^N y_{i}\right)^2}{n} \ge 0,$

where $\displaystyle n=\sum_{i=1}^N n_{i},\ \ n_{i} \ge 0$, and $\displaystyle y_{i}$ is a real quantity (can be both negative or positive)?

It appears that the inequality is valid (even used random numbers), but can't see how to prove it.

This inequality gets really messy really quickly. So I will prove the special case $\displaystyle N=2$. I am sure you can use the same method to generalize it to more terms but it just is so messy.
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First, I think you meant $\displaystyle n_i >0$.

Thus, we have to show:
$\displaystyle \frac{y_1^2}{n_1}+\frac{y_2^2}{n_2} \geq \frac{(y_1+y_2)^2}{n_1+n_2}$

Rewrite as,
$\displaystyle \frac{y_1^2}{n_1}+\frac{y_2^2}{n_2} \geq \frac{y_1^2}{n_1+n_2}+\frac{y_2^2}{n_1+n_2}+ 2\cdot \frac{y_1y_2}{n_1+n_2}$

Multiply by $\displaystyle n_1n_2(n_1+n_2)>0$:

$\displaystyle n_2(n_1+n_2)y_1^2+n_1(n_1+n_2)y_2^2 \geq y_1^2n_1n_2+y_2^2n_1n_2+2y_1y_2n_1n_2$

Open and cancel,

$\displaystyle n_2^2y_2^2 + n_1^2y_2^2 \geq 2y_1y_2n_1n_2$

This is the AM-GM inequality.
Which is true.

(Or you can write $\displaystyle (n_1y_1 - n_2y_2)^2 \geq 0$).