The basic arithmetic meangeometric mean (AMGM) 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 a^{2}+b^{2} = c^{2}, 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
2CT^2 = (AC^2 + BC^2)  AT^2  TB^2
= AB^2  AT^2  TB^2
= (AT + TB)^2  AT^2  TB^2
= 2ATTB
= 2xy
Therefore,
sqrt(xy) = CT <= AB/2 = (AT + TB)/2 = (x + y)/2

Posted by Bractals
on 20100305 18:11:49 