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!

Question

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!

anonymous · Answer

$$\frac{ a + b }{ 2 } \ge \sqrt{ab}$$

watchmath · Answer

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 · Answer

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)?

anonymous · Answer

I should probably say that no calculus is required :)

homeworksucks · Answer

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

watchmath · Answer

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 · Answer

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 · Answer

The proof is as I said in the first post

anonymous · Answer

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