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.
 A: How about the following solution?

A: What about something like this:
$\hspace{80pt}$
The two right triangles have common base, which is also the diameter
of the red circle. The endpoints of the black diameter are also
the foci of the ellipses, and their size is determined by the tip of the respective triangle.
Observe, that the minor radius of the blue ellipse is equal to the radius of the red circle, thus the circle is inscribed into the blue ellipse.
Therefore, as the tip of the green triangle is contained within
the blue ellipse, its perimeter has to be smaller or equal the perimeter of the blue triangle, which implies the inequality between the means.
$$\sqrt{2a^2+2b^2}\geq a+b.$$
More visually, the green ellipse has the same focal points as the blue one, but intersects the red circle in 4 points (the tip and 3 reflections), while the blue does only in two. Hence the radii of the green ellipse has to be smaller or equal the radii of the blue ellipse, which also implies the inequality between the means.
I hope this helps $\ddot\smile$
A: 
$\qquad\qquad 2 \left(\;\color{blue}{a}^2 + \color{red}{b}^2\;\right) \;\geq\; \left(\;\color{blue}{a}+\color{red}{b}\;\right)^2$
A: $$\sqrt{\frac{a^2 + b^2}{2}} \ge \frac{a + b}{2} $$
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).
