# Should be pretty simple proof

1. Sep 3, 2009

### Newtime

...but I feel like I did something wrong. Also, this was a problem in my Analysis book, hence my posting it here, although it doesn't explicitly deal with analysis.

Prove that root(ab)=(a+b)/2 implies a=b. Assume 0$$\leq$$a$$\leq$$b. To prove the converse is true was another problem and was easy but anyway here's my work:

Proof.

Assume the contrary; that given root(ab)=(a+b)/2, a$$\neq$$b.

By the first multiplicative identity, 2*root(ab)=(a+b).

Squaring both sides: 4ab=a2+2ab+b2

By the first multiplicative identity and algebra, a2-2ab+b2=0.

Factor: (a-b)(a-b)=0.

Since a$$\neq$$b by assumptions, a-b$$\neq$$0 thus we can divide both sides by a-b: (a-b)=0. This implies a=b, which contradicts the original assumptions. Thus root(ab)=(a+b)/2 implies a=b.

qed

So is this correct? If not, where did I go wrong? I just have this feeling that it's a little...off...somewhere but I don't know how or where. Thanks in advance for the help, this has been bothering me for a bit now.

2. Sep 3, 2009

### snipez90

This looks fine and is the equality condition in the 2-variable AM-GM inequality. Anyways, it is an exercise in basic analysis to show that if a and b are real numbers and ab = 0, then either a or b is 0. This might be the reason why you feel the proof is a bit weird, but it looks correct to me.

3. Sep 8, 2009

### millern64

it doesn't look like you need to prove by contradiction. just following the steps shows explicitly that a must equal b. then shouldn't need to assume 0=<a=<b (which forces a and b to be positive, both can be negative as well) ...otherwise it looks good.

4. Sep 9, 2009

### rzaidan

This is the first time that I participate this site and I have the following proof for your problem:
(a-b)^2=a^2-2ab+b^2=a^2+2ab+b^2-4ab=(a+b)^2-4ab
=4[((a+b)/2)^2-ab]
=4([(a+b)/2)-sqrt{ab})([(a+b)[(a+b)/2)+sqrt{ab})
=0 since (a+b)/2)=sqrt{ab}
therefore (a-b)^2=0 implies a-b=0 and a=b Q.E.D
Best Wishes

5. Sep 9, 2009

### arildno

Why is (a+b)/2 equal to sqrt{ab}?

The geometric mean of two numbers are hardly ever equal to the arithmetic mean..

6. Sep 9, 2009

### HallsofIvy

Staff Emeritus
Arildno, read the original post! The problem was to prove "If (a+b)/2= sqrt(ab), then a= b".

7. Sep 9, 2009

### lurflurf

let h=(b-a)/2
so 0<=h
sqrt(ab)=(a+b)/2
sqrt(a^2+2ah)=a+h

8. Sep 10, 2009

### rzaidan

Dear arildno
the problem was to prove that
"If (a+b)/2= sqrt(ab), then a= b".So this is given
Best Regards

9. Sep 13, 2009