Check my inequality proof please

  • Thread starter skeeterrr
  • Start date
  • #1
14
0

Homework Statement



Prove for all real numbers x and y that [tex]2xy =< x^2 + y^2[/tex]

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?
 

Answers and Replies

  • #2
186
0
i think that it looks fine
 
  • #3
VietDao29
Homework Helper
1,423
3
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.
 

Related Threads on Check my inequality proof please

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
5
Views
1K
Replies
2
Views
827
Replies
2
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
11
Views
1K
Top