Three sticks, one triangle

Idea. The triangle fails if and only if one side is at least the sum of the other two. These failures are disjoint, so their probabilities add, and each one is a simple volume.

Step 1: failure events. No triangle exists when $X\ge Y+Z$, or $Y\ge X+Z$, or $Z\ge X+Y$. Two of these cannot hold together (adding them would force a length to be $0$, which has probability $0$). So $$\mathbb{P}(\text{no triangle})=3\mathbb{P}(Z\ge X+Y).$$

Step 2: one volume. For fixed $Z=z$, the set $\{x+y\le z\}$ in the unit square is a right triangle with area $\frac{z^2}{2}$ (since $z\le 1$). Hence $$\mathbb{P}(Z\ge X+Y)={\int }_0^1\frac{z^2}{2}dz=\frac{1}{6}.$$

Step 3. $$\mathbb{P}(\text{triangle})=1-3\cdot \frac{1}{6}=\frac{1}{2}.$$

Compare. If instead a single stick of length $1$ is broken at two uniform points, the three pieces form a triangle with probability $\frac{1}{4}$. The constraint that the pieces add up to $1$ makes a long piece more likely.