Posted in MO since it is unanswered in MSE.
Let $(a,b,c)$ be the side of a triangle. In its most general linear form, the triangle inequality can be expressed as: Does $ax + by \ge c$ for fixed $x,y \ge 0$ hold? Trivially the inequality holds if both $x$ and $y$ are $\ge 1$; however if one of or both of $x$ and $y$ is non-negative and $\le 1$ then $ax+y \ge c$ is not necessarily true. Assuming that vertices of a triangle are uniformly random on a circle we can ask the probability $P(ax+by \ge c)$. In this question we found a closed form for the probability $P(a+b \ge cx)$, $x \ge 1$. This question is an attempt at generalization of this result.
Experimental data show that if $0 \le x,y \le 1$ then, $$ P(ax + by \ge c) = \frac{4}{\pi^2}\chi_2(x) + \frac{4}{\pi^2}\chi_2(y) \tag 1 $$ and if $0 \le x \le 1 \le y$ then, $$ P(ax + by \ge c) = 1 + \frac{4}{\pi^2}\chi_2\left(\frac{x}{y}\right) - \frac{4}{\pi^2}\chi_2\left(\frac{1}{y}\right) \tag 2 $$
where $\displaystyle \chi_2(x) = \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)^2}, |x| \le 1$ is the Legendre Chi function.
Question 1: Is $(1)$ true?
Question 2: Is $(2)$ true?
