# Help with the proof

1. May 24, 2013

### woundedtiger4

If a and b are real numbers, then a^2 + b^2 >= 2 a b
Proof:
if a>=0 and b>=0
then a-b>=0
(a-b)^2>=0
a^2 + b^2 -2ab >=0
a^2 + b^2 >= 2ab
Q.E.D.

someone has proved more or less similar question, but at

(a + b)^2 > 0
a^2 + 2ab + b^2 > 0
a^2 + b^2 ≥ 2ab

when 2ab goes from L.H.S. to R.H.S then it should be -2ab , am I correct?

2. May 24, 2013

### Fredrik

Staff Emeritus
Yes, and that's why you should use $(a-b)^2$ instead of $(a+b)^2$ to prove this result.

You should remove the first two lines from the first proof. Do you see why?

3. May 24, 2013

### woundedtiger4

Because a and b could be equal, right?

If I am correct then it should be:

Proof:
If a-b>=0
then (a-b)^2 >= 0
a^2 + b^2 - 2ab >= 0
implies a^2 + b^2 >= 2ab
Q.E.D.

But if a and b could be equal to each other then ">=" justifies that so the proof in my original post remains correct, isn't?

4. May 24, 2013

### HallsofIvy

This is not true. Given that a and b are arbitrary real numbers, a- b can be any number, positive, negative, or 0.
However, since a square of a real number is never negative, the next step is correct:

If a and b are not allowed to be equal, yes. otherwise (a+ b)^2>= 0 as you had before.

Yes, but since a and b can be any real numbers, either 2ab< 0, in which case, since a^2+ b^2 is not negative, it is a trivial statement, or 2ab> 0 so that the sign doesn't matter.

5. May 24, 2013

### woundedtiger4

Proof:
If a-b >=0
then (a-b)^2 >=0
a^2 +b^2 -2ab >=0
a^2 + b^2 >=2ab
Q.E.D.

What do you say?

6. May 24, 2013

### Office_Shredder

Staff Emeritus

You seem to be stuck on this idea of a-b ≥ 0 that is completely unnecessary. Compare your proof to the following:

If the sky is blue
then (a-b)^2 >=0
a^2 +b^2 -2ab >=0
a^2 + b^2 >=2ab

7. May 24, 2013

### woundedtiger4

Ohh got it. It says "a and b are real numbers".

Proof:
If (a-b)^2 >=0
then a^2 + b^2 >=0
Q.E.D.

Right?

8. May 24, 2013

### Fredrik

Staff Emeritus
No, the problem is that if the proof relies on the assumption that $a,b\geq 0$, then the result will only be valid for such values of $a$ and $b$

Don't forget that the statement you want to prove is that "For all real numbers $a$ and $b$, we have $a^2+b^2\geq 2ab$. It's for all real numbers, not just for the non-negative ones.

What makes you think that $(a-b)^2$ may be less than 0, or that it needs to be greater than or equal to zero to ensure that $a^2+b^2\geq 0$?

9. May 25, 2013

### woundedtiger4

Proof:
If a,b>=0
then (a-b)^2 >=0
a^2 +b^2 -2ab >= 0
a^2 + b^2 >= 2ab
QED

I don't understand that how come (a+b)^2 >=0 can lead us to prove that a^2 + b^2 >= 2ab because (a+b)^2 >= 0 will lead us to the resut a^2 +b^2 >= -2ab , please explain to me in a simple language becaueee I am a beginner.

PS. I want to prove this thm by using direct proof's if & then technique.

10. May 25, 2013

### micromass

Can you give me some real numbers such that $(a-b)^2<0$?

11. May 25, 2013

### Fredrik

Staff Emeritus
I didn't express myself very clearly here. I meant that if you assume that $a,b\geq 0$ and then find that $a^2+b^2\geq 2ab$, you will only have proved that $a^2+b^2\geq 2ab$ for all non-negative real numbers $a$ and $b$.

But your goal was to prove that $a^2+b^2\geq 2ab$ for all real numbers $a$ and $b$.