Fun Challenge
So here is the challenge. Prove that (a + b)/2 (the arithmetic mean) is greater than or equal to sqrt(ab) (the geometric mean) for any real numbers a and b!

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- anonymous

- katieb

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- anonymous

\[\frac{ a + b }{ 2 } \ge \sqrt{ab}\]

- watchmath

Squaring both sides and rearranging we have \((a-b)^2\geq 0\) which is always true. BTW you need the condition that \(a,b\) are non negative though!!

- Homeworksucks

5+6/2=11
sqrt(30)= 5.47722557505
11 is greater than 5.4
1+1.00001/2= 1.00005
1*1.00001=1.00001
sqrt(1.00001)= 1.00000499999
I haven't studied calculus yet but isn't something like as a+b approaches 2 at the very limit, it should be equal to sqrt(ab)?

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- anonymous

I should probably say that no calculus is required :)

- Homeworksucks

Dammit, that probably wasn't the kind of proof you were looking for wasn't it?

- watchmath

a=-2 b=-2. Arithmetic mean is -2 , but geometric mean is 2. So AM < GM in this case. You need a and b to be positive

- anonymous

Good point! That's what I get for working out of a textbook, huh? Great counterexample. Ok let's assume a and b are positive then.

- watchmath

The proof is as I said in the first post

- anonymous

You were not very explicit but you got the same result I did. Magic?

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