Školjka

Let $ABC$ be a triangle. The points $K$, $L$, and $M$ lie on the segments $BC$, $CA$, and $AB$, respectively, such that the lines $AK$, $BL$, and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM$, $BMK$, and $CKL$ whose inradii sum up to at least the inradius of the triangle $ABC$. \begin{flushright}\emph{(Estonia)}\end{flushright}