# $\sum{x} \ge \sum{y}$ implies $\sum{x^2} \ge \sum{y^2}$

For any two sets of n non-negative integers, $x_i$ and $y_i$, is the above true, and how can one prove it? It seems like it ought to be true. I think induction is probably the best way, so I probably just need help to prove that if $x_1+x_2 \ge y_1 + y_2$ then $x_1^2 + x_2^2 \ge y_1^2 + y_2^2$ which I can't seem to manage.

• note that $4+4>6+1$ but $4^2+4^2<6^2 +1^2$. – Rolf Hoyer Oct 5 '16 at 20:20
• If you have ever taken a class on statistics you may recall the key measurements of a sample: the average and the variance. Here you are given two sample with one having a higher average than the other. But there's no information about the variance. And variance can be written in terms of the average and the sum of squares. So if you pick two samples with nearly the same averages, but the lower average sample having a much higher variance, you get ... counterexamples. – Jyrki Lahtonen Oct 5 '16 at 20:32

## 1 Answer

It's wrong.

$2+2+2\geq1+1+4$, but $2^2+2^2+2^2\geq1^2+1^2+4^2$ is wrong.

• Lol ok I was an idiot – Bob Oct 5 '16 at 20:28
• @Bob You're not an idiot. Intuition is often wrong, and that's how you learn. If you feel like this answered your question, you should consider accepting it (clicking the tick). – Eff Oct 5 '16 at 20:33