Geometric proof of $QM \ge AM$ Prove by geometric reasoning that:
$$\sqrt{\frac{a^2 + b^2}{2}} \ge \frac{a + b}{2}$$
The proof should be different than one well known from Wikipedia:

DISCLAIMER: I think I devised such proof (that is not published anywhere to my knowledge), but I believe it is too complex (and not that beautiful). It involves some ideas from another question. I may post it in next few days as an answer, but I am hoping someone will offer better, more beautiful, proof by then.

$$\sqrt{\frac{a^2 + b^2}{2}} \ge \frac{a + b}{2} $$
To prove, square and subtract numerator and denominator separately getting
$$\frac{a^2 + b^2}{2} - {(\frac{a + b}{2}})^2 = a b > 0.$$
(In a geometrical reasoning an algebraic process is inherent.. even in the Pythagoras thm).
