Probability that the reciprocal triangle inequality $\frac{1}{a} + \frac{1}{b} \ge \frac{1}{c}$ holds

Question

Let $(a,b,c)$ be the sides of a triangle inscribed inside a unit circle such that the vertices of the triangle are distributed uniformly on the circumference. The regular triangle inequality states that the sum of any two sides is greater than the third side. But what happens if we take the sum of the reciprocal of any two sides? Is it greater than the third side? It turns out that the reciprocal triangle inequality $\frac{1}{a} + \frac{1}{b} \ge \frac{1}{c}$ is not true in general however, experimental data shows an interesting observation that the probability

$$ P\left(\frac{1}{a} + \frac{1}{b} \ge \frac{1}{c}\right) = \frac{4}{5} $$

Can this be proved or disproved? Note that this is equivalent proving or disproving

$$ P\left(\frac{1}{\sin x} + \frac{1}{\sin y} \ge \frac{1}{|\sin (x+y)|} \right) = \frac{4}{5} $$

where $0 \le x,y \le \pi$.

Related question: Probability that the geometric mean of any two sides of a triangle is greater than the third side is $\displaystyle \frac{2}{5}$.

Answer 1

Assume that the circle is centred at the origin, and the vertices of the triangle are:
$A(\cos(-2Y),\sin(-2Y))$ where $0\le Y\le\pi$
$B(\cos(2X),\sin(2X))$ where $0\le X\le\pi$
$C(1,0)$

Let:
$a=BC=2\sin X$
$b=AC=2\sin Y$
$c=AB=\left|2\sin\left(\frac{2\pi-2X-2Y}{2}\right)\right|=|2\sin(X+Y)|$

$P\left[\frac1a+\frac1b\ge\frac1c\right]=1-P\left[\frac1a+\frac1b<\frac1c\right]$

$P\left[\frac1a+\frac1b<\frac1c\right]=P\left[\frac{1}{\sin X}+\frac{1}{\sin Y}<\frac{1}{|\sin(X+Y)|}\right]$

This last probability is the ratio of the area of the shaded region to the area of the square in the graph below.

Rotate these regions $45^\circ$ clockwise about the origin, then shrink by factor $\frac{1}{\sqrt2}$, by letting $X=x-y$ and $Y=x+y$.

Using symmetry, we only need to consider the left half of the blue "diamond". Note that in the left half, $0<x<\pi/2$, so $|\sin(2x)|=\sin(2x)$.

$P\left[\frac{1}{\sin X}+\frac{1}{\sin Y}<\frac{1}{|\sin(X+Y)|}\right]$
$=P\left[\frac{1}{\sin(x-y)}+\frac{1}{\sin(x+y)}<\frac{1}{\sin(2x)}\right]$

Now we want to express the inequality as $-f(x)<y<f(x)$ for some $f(x)$.

$=P\left[\sin(2x)(\sin(x+y)+\sin(x-y))<\sin(x-y)\sin(x+y)\right]$
$=P\left[2\sin(2x)(\sin x)(\cos y)<(\sin^2x)(\cos^2y)-(\cos^2x)(\sin^2y)\right]$
$=P\left[2\sin(2x)(\sin x)(\cos y)<(\sin^2x)(\cos^2y)-(\cos^2x)(1-\cos^2y)\right]$
$=P\left[\cos^2y-2\sin(2x)(\sin x)(\cos y)-\cos^2x>0\right]$

Solving the quadratic in $\cos y$ gives

$=P\left[\cos y>(\cos x)\left(2\sin^2x+\sqrt{1+4\sin^4x}\right)\right]$
$=P\left[-f(x)<y<f(x)\right]$

where $\color{red}{f(x)=\arccos\left((\cos x)\left(2\sin^2x+\sqrt{1+4\sin^4x}\right)\right)}$

Note that $f(\arccos(1/4))=0$ and $f(\pi/2)=\pi/2$.

So we have $P\left[\frac1a+\frac1b\ge\frac1c\right]=1-\dfrac{\int_{\arccos(1/4)}^{\pi/2}f(x)\mathrm dx}{\frac12 \left(\frac{\pi}{2}\right)^2}$

Here it is shown that $\int_{\arccos(1/4)}^{\pi/2}f(x)\mathrm dx=\dfrac{\pi^2}{40}$.

$\therefore P\left[\frac1a+\frac1b\ge\frac1c\right]=1-\dfrac15=\dfrac45$.