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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
 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
 one year ago

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

watchmathBest ResponseYou've already chosen the best response.1
Squaring 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!!
 one year ago

HomeworksucksBest ResponseYou've already chosen the best response.0
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)?
 one year ago

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

HomeworksucksBest ResponseYou've already chosen the best response.0
Dammit, that probably wasn't the kind of proof you were looking for wasn't it?
 one year ago

watchmathBest ResponseYou've already chosen the best response.1
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
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

LogicalAppleBest ResponseYou've already chosen the best response.0
You were not very explicit but you got the same result I did. Magic?
 one year ago
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