anonymous
  • anonymous
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!
Mathematics
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anonymous
  • anonymous
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!
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ a + b }{ 2 } \ge \sqrt{ab}\]
watchmath
  • 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
  • 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
  • anonymous
I should probably say that no calculus is required :)
Homeworksucks
  • Homeworksucks
Dammit, that probably wasn't the kind of proof you were looking for wasn't it?
watchmath
  • 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
  • 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
  • watchmath
The proof is as I said in the first post
anonymous
  • anonymous
You were not very explicit but you got the same result I did. Magic?

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