Triangle inequality

Idea

Work with the complement: a triple fails the triangle inequality if and only if one side is at least as large as the sum of the other two. The three failure events turn out to be mutually exclusive, so the failure probability is simply $3P(A)$ for any one of them.

Setup

Let $X, Y, Z \overset{\text{iid}}{\sim} \text{Uniform}[0,1]$. Define the failure events

Mutual Exclusivity of $A$, $B$, $C$

Suppose $A \cap B$ occurs: $X+Y \le Z$ and $X+Z \le Y$. Adding these inequalities gives $2X + Y + Z \le Y + Z$, hence $X \le 0$. Since $X \ge 0$ a.s., this forces $X = 0$, an event of probability zero. By symmetry, every pairwise intersection has measure zero, so $A$, $B$, $C$ are mutually exclusive a.s. Combined with symmetry of the joint distribution,

Computing $P(A)$

Condition on $Z = z$. The event $\{X + Y \le z\}$ within $[0,1]^2$ traces the right triangle $\{x \ge 0,\, y \ge 0,\, x+y \le z\}$, which has area $z^2/2$. Therefore

Final Answer