1. Sep 18, 2009

skeeterrr

1. The problem statement, all variables and given/known data

Prove for all real numbers x and y that $$2xy =< x^2 + y^2$$

2. Relevant equations

3. The attempt at a solution

Well, since this is a problem regarding proof, I thought I would start with a contradictory statement like:

2xy >= x^2+y^2

0 >= x^2-2xy+y^2

0 >= (x-y)^2

Since (x-y)^2 is either a positive integer and a zero,

0 =< (x-y)^2

0 =< x^2-2xy+y^2

2xy =< x^2+y^2

Well that's all I can think of... can anyone point out any mistakes or anything? Is there more to it than just this?

2. Sep 18, 2009

nietzsche

i think that it looks fine

3. Sep 19, 2009

VietDao29

What you've done is not a Proof by Contradiction.

The bottom part of your post is indeed qualified as a direct proof. No need for Contradiction. :)

And btw, the negation of 2xy <= x2 + y2 is not

Instead, it should be: 2xy > x2 + y2.