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.

Let ATB be a line segment with

  x = |AT|  and  y = |TB|.

Construct a semicircle with AB as diameter.

Construct line segment CT with C on the 
semicircle and CT perpendicular to AB.

Triangle ABC is clearly a right triangle
with |CT| <= |AB|/2 ( with equality when
x = y ).

   |CT|^2 = |AC|^2 - |AT|^2

   |CT|^2 = |BC|^2 - |TB|^2

  2|CT|^2 = (|AC|^2 + |BC|^2) - |AT|^2 - |TB|^2

          = |AB|^2 - |AT|^2 - |TB|^2

          = (|AT| + |TB|)^2 - |AT|^2 - |TB|^2

          = 2|AT||TB|

          = 2xy

Therefore,

  sqrt(xy) = |CT| <= |AB|/2 = (|AT| + |TB|)/2 = (x + y)/2

Posted by Bractals on 2010-03-05 18:11:49