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).