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 one year 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!
 one year 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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LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ a + b }{ 2 } \ge \sqrt{ab}\]

watchmath
 one year 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
 one year 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)?

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I should probably say that no calculus is required :)

Homeworksucks
 one year 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
 one year 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

LogicalApple
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1The proof is as I said in the first post

LogicalApple
 one year 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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