How robust is the AM-GM inequality? Suppose we pick $\lambda$ with a constant probability distribution in the interval $[0,1]$ and $x>0$ and $y>0$, also with uniform distribution in the first quadrant up to distance $R$ from the origin (a fourth of circle). Then what's the probability that
\[
\lambda x+(1-\lambda)y\geq \sqrt{xy}?
\]
My attempt at answering this is as follows:
In order for this to happen, supposing that $x>y$
\[
\lambda \geq \frac{\sqrt{xy}-y}{x-y}=\frac{\sqrt{x/y}-1}{x/y-1}=\frac{1}{1+\sqrt{x/y}}
\]
Going to polar coordinates, with $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have
\[
\lambda\geq \frac{1}{1+\sqrt{\cot(\theta)}}
\]
For this value of $\theta$, the probability that the "weighted AM-GM inequality" is valid
 is 
\[
p(\theta)=1-\lambda(\theta)=\frac{\sqrt{\cot(\theta)}}{1+\sqrt{\cot(\theta)}}
=\frac{1}{1+\sqrt{\tan(\theta)}}
\]
When we average on the angle $\theta$ we get the probability of picking
random numbers that satisfy the W-AM-GM inequality. This results in
the probability
\[
\frac{4}{\pi}\int_0^{\pi/4}\frac{d\theta}{1+\sqrt{\tan(\theta)}}
\]
Is this correct? 
How to generalize this to $N$ points? i.e.
What's the probability of picking random points ${0\leq p_i\leq 1}$ and $x_i>0$ such that
\[
\sum_{i=1}^Np_ix_i\geq(x_1x_2\dots x_N)^{1/N},
\]
where $\sum_{i=1}^Np_i=1$

I believe your expression is correct for the case of N=2 except for the bounds of integration. When you divide by "X-Y" and don't flip the inequality you are assuming X-Y is positive or X>Y so you only need to integrate over the lower diagonal of the first quadrant i.e. theta=(0,pi/4).
