Denote
$$a = \frac{BK}{KC}, \quad b = \frac{CL}{LA}, \quad c = \frac{AM}{MB}$$
By Ceva's theorem, $abc = 1$, so we may, without loss of generality, assume that $a \geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than $1$. Therefore at least one of the pairs $(a, b)$, $(b, c)$ has its first component not less than $1$ and the second one not greater than $1$. Without loss of generality, assume that $1 \leqslant a$ and $b \leqslant 1$.

Therefore, we obtain $bc \leqslant 1$ and $1 \leqslant ca$, or equivalently
$$\frac{AM}{MB} \leqslant \frac{LA}{CL} \quad \text{and} \quad \frac{MB}{AM} \leqslant \frac{BK}{KC}$$
The first inequality implies that the line passing through $M$ and parallel to $BC$ intersects the segment $AL$ at a point $X$ (see Figure 1). Therefore the inradius of the triangle $ALM$ is not less than the inradius $r_1$ of triangle $AMX$.

Similarly, the line passing through $M$ and parallel to $AC$ intersects the segment $BK$ at a point $Y$, so the inradius of the triangle $BMK$ is not less than the inradius $r_2$ of the triangle $BMY$. Thus, to complete our solution, it is enough to show that $r_1 + r_2 \geqslant r$, where $r$ is the inradius of the triangle $ABC$. We prove that in fact $r_1 + r_2 = r$.

Figure 1

Since $MX \parallel BC$, the dilation with centre $A$ that takes $M$ to $B$ takes the incircle of the triangle $AMX$ to the incircle of the triangle $ABC$. Therefore
$$\frac{r_1}{r} = \frac{AM}{AB}, \quad \text{and similarly} \quad \frac{r_2}{r} = \frac{MB}{AB}$$
Adding these equalities gives $r_1 + r_2 = r$, as required.