Proving 2xy =< x^2 + y^2 for all real numbers x and y: A Contradictory Approach

[skeeterrr]

Homework Statement

Prove for all real numbers x and y that 2xy =&lt; x^2 + y^2

Homework Equations

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?

 

[nietzsche]

i think that it looks fine

 

[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. :)

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?

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

2xy >= x^2+y^2

Instead, it should be: 2xy > x² + y².