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 Pythagoras knew it!! (Posted on 2010-03-05)
The basic arithmetic mean-geometric mean (AM-GM) inequality states simply that if x and y are nonnegative real numbers, then. (x+y)/2 ≥ √(x*y), with equality if and only if x = y.
There are various proofs for this theorem (for any number of values), inter alia Polya, Cauchy, by induction etc.
Now derive your proof directly from Pythagoras' formula a2+b2 = c2, a ≠ b.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 my attempt | Comment 1 of 5
a^2 + b^2 = c^2
c^2 - a^2 = b^2 >= 0
c^2 - a^2 >= 0
(c-a) * (c+a) >= 0
(c-a)^2 * (c+a)^2 >= 0
(c-a)^2 >= 0
c^2 - 2ca + a^2 >= 0
c^2 + 2ca + a^2 >= 4ca
(c+a)^2 >= 4ca
c+a >= 2*sqrt(ca)  of course this step holds if a,c are non-negative
(c+a)/2 >= sqrt(ca)
thus for two non-negative real numbers their arithmetic mean is greater than their geometric mean.

 Posted by Daniel on 2010-03-05 17:10:36

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