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
sqrt(xy) = |CT| <= |AB|/2 = (|AT| + |TB|)/2 = (x + y)/2
Posted by Bractals
on 2010-03-05 18:11:49