The Problems of Ukrainian Authors

Denote the sides of triangle by $a$, $b$, $c$. By the property of angular bisectors,

$AD=\frac{ch}{a+c}$, $AE=\frac{ch}{a+b}$ (pic.30), thus it is enough to prove: $\frac{ch}{a+c}+\frac{ch}{a+b}\le a$.

After simple transformations, we have: $(a^2-bc)(a+b+c)\ge 0$, and so it is enough to prove: $a^2-bc\ge 0$.

By the cosine theorem, $a^2=b^2+c^2-2bc\cos A\ge b^2+c^2-2bc\cdot \frac{1}{2}\ge 2bc-bc=bc$, which required.