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anonymous
 4 years ago
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
 4 years ago
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
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ a + b }{ 2 } \ge \sqrt{ab}\]

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.1Squaring both sides and rearranging we have \((ab)^2\geq 0\) which is always true. BTW you need the condition that \(a,b\) are non negative though!!

Homeworksucks
 4 years ago
Best ResponseYou've already chosen the best response.05+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
 4 years ago
Best ResponseYou've already chosen the best response.0I should probably say that no calculus is required :)

Homeworksucks
 4 years ago
Best ResponseYou've already chosen the best response.0Dammit, that probably wasn't the kind of proof you were looking for wasn't it?

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.1a=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
 4 years ago
Best ResponseYou've already chosen the best response.0Good 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
 4 years ago
Best ResponseYou've already chosen the best response.1The proof is as I said in the first post

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You were not very explicit but you got the same result I did. Magic?
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