# Homework Help: Prove 2ab<= a^2+b^2 using order axioms

1. Oct 5, 2012

### whyme1010

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

prove 2ab<= a^2+b^2 using order axioms
Of course use of other basic axioms for real numbers are also okay.

2. Relevant equations
axioms for set of real numbers.

3. The attempt at a solution
The easy way to do this would be just subtract 2ab from both sides, factor, and see that (a-b)^2 is greater or equal to zero.

But we have to use the basic axioms. So I tried constructing a^2+b^2-2ab by multiplying (a-b)(a-b) using the axioms. I'm not sure I was rigorous enough. Also I'm not sure whether I can just say a square of a real number is greater than zero. Though it is easy to prove.

2. Oct 5, 2012

### happysauce

You can use only the order axioms and the field axioms. I don't think you can use squared numbers are greater than zero.

3. Oct 5, 2012

### whyme1010

well then I can just prove that a^2 is always greater or equal to zero.

a>0
Then a(a)>(a)0
a^3>0

a<0
a+(-a)<0+(-a)
0<-a
-a(0)<(-a)(-a)
0<(-1)(-1)(a)(a)
0<a

a=0
a(a)=0(a)
a^2=0

So a^2 >=0

4. Oct 8, 2012

### happysauce

Hmmm I guess so. did you prove that (-x)(-y)=-(xy)?

5. Oct 8, 2012

### whyme1010

yeah. I also solved this by using three lemmas. =) Don't know how to close the thread though.

6. Oct 8, 2012

### Mentallic

I sure hope he didn't.

7. Oct 8, 2012

### happysauce

woops I mean (x)(-y) = -(xy)