Solution:
We must show that $r$, $r_{1}$, and $r_{2}$ satisfy the triangle inequality, i.e. that the sum of any two of them exceeds the third. Clearly $r$ is the largest of the three, so we need only verify that $r_{1} + r_{2} > r$.

Let $K$ and $s$ be the area and semiperimeter of triangle $ABC$. Similarly define $K_{1}$, $K_{2}$, $s_{1}$, and $s_{2}$. Observe that $s$ is larger than $s_{1}$ or $s_{2}$ and that $K_{1} + K_{2} = K$. While these facts are almost trivial to verify, they must be stated.

Then $r = K / s$, $r_{1} = K_{1} / s_{1}$, and $r_{2} = K_{2} / s_{2}$, so
$$r_{1} + r_{2} = \frac{K_{1}}{s_{1}} + \frac{K_{2}}{s_{2}} > \frac{K_{1}}{s} + \frac{K_{2}}{s} = \frac{K}{s} = r.$$