, 2015

Denote
$$a = \frac{BK}{KC}, \quad b = \frac{CL}{LA}, \quad c = \frac{AM}{MB}.$$
By Ceva's theorem we know $abc = 1$. Hence without loss of generality we may assume $a \ge 1$. Then at least one of $b, c$ is not greater than 1. Therefore, among the two pairs $(a, b), (b, c)$, at least one pair has its first number not less than 1 and its second number not greater than 1. Without loss of generality, assume further that $1 \le a$ and $b \le 1$.
From this we obtain $bc \le 1$ and $1 \le ca$, that is,
$$\frac{AM}{MB} \le \frac{LA}{CL} \text{ and } \frac{MB}{AM} \le \frac{BK}{KC}.$$
The first inequality above tells us that: the line through $M$ parallel to $BC$ meets segment AL at a point $X$. Therefore, the inradius of triangle $ALM$ is not less than the inradius $r_1$ of triangle $AMX$.
Similarly, the second inequality indicates that: the line through $M$ parallel to $AC$ meets segment BK at a point $Y$, so the inradius of triangle $BMK$ is not less than the inradius $r_2$ of triangle $BMY$. To complete the proof, it suffices to show that $r_1 + r_2 \ge r$, where $r$ is the inradius of triangle $ABC$. In fact, we will prove that: $r_1 + r_2 = r$.

Since $MX \parallel BC$, the homothety centered at $A$ that sends $M$ to $B$ will send the incircle of triangle $AMX$ to the incircle of triangle $ABC$. Hence we have
$$\frac{r_1}{r} = \frac{AM}{AB}.$$
Similarly we obtain
$$\frac{r_2}{r} = \frac{MB}{AB}.$$
Adding these two equations gives the desired result.