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LogicalApple

  • 3 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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  1. LogicalApple
    • 3 years ago
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    \[\frac{ a + b }{ 2 } \ge \sqrt{ab}\]

  2. watchmath
    • 3 years ago
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    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!!

  3. Homeworksucks
    • 3 years ago
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    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)?

  4. LogicalApple
    • 3 years ago
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    I should probably say that no calculus is required :)

  5. Homeworksucks
    • 3 years ago
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    Dammit, that probably wasn't the kind of proof you were looking for wasn't it?

  6. watchmath
    • 3 years ago
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    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

  7. LogicalApple
    • 3 years ago
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    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.

  8. watchmath
    • 3 years ago
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    The proof is as I said in the first post

  9. LogicalApple
    • 3 years ago
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    You were not very explicit but you got the same result I did. Magic?

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