Iranian Mathematical Olympiad

Solution:

Denote by $I$ the intersection point of $AD$ and $BE$, so $I$ is the incenter of triangle $ABC$. Suppose that $\alpha =\frac{\angle BAC}{2}$. We have $$\angle IBD=90^{\circ }-\angle IBA=90^{\circ }-\frac{\angle CBA}{2}=90^{\circ }-30^{\circ }=60^{\circ },$$ $$\angle CEB=90^{\circ }-\angle CBE=90^{\circ }-\frac{\angle CBA}{2}=90^{\circ }-30^{\circ }=60^{\circ }.$$ Thus in triangles $BEC$ and $BED$ we have $\angle CEB=\angle EBD=60^{\circ }$ and $BE$ is side of both of them. Since $\angle BEC=30^{\circ }$ it suffices to prove $BD<CE$ but we have $\frac{BD}{AB}=\tan \alpha$ and $\frac{EC}{BC}=\tan 30^{\circ }$, So $$BD\le CE\iff AB\tan \alpha \le BC\tan 30^{\circ }\iff \frac{AB}{BC}\tan \alpha \le \tan 30^{\circ }$$ $$\iff \tan \alpha \frac{\sin (120^{\circ }-2\alpha )}{\sin (2\alpha )}\le \tan 30^{\circ }(\text{Law of Sines})$$ $$\iff \frac{\sin \alpha }{\cos \alpha }\frac{\sin (120^{\circ }-2\alpha )}{2\sin \alpha \cos \alpha }\le \tan 30^{\circ }$$ $$\iff \sin (120^{\circ }-2\alpha )\le 2{\cos }^2\alpha \cdot \tan 30^{\circ }$$

$$\begin{array}{rl}& \iff \sin 120^{\circ }\cos (2\alpha )-\sin 120^{\circ }\sin (2\alpha )\le (1+\cos (2\alpha ))\tan 30^{\circ }\\ & \iff (\sin 120^{\circ }-\tan 30^{\circ })\cos (2\alpha )-\cos 120^{\circ }\sin (2\alpha )\le \tan 30^{\circ }.\end{array}$$ We know by Cauchy-Schwarz inequality that $$\begin{array}{rl}(LHS{)}^2& \le ((\sin 120^{\circ }-\tan 30^{\circ }{)}^2+(\cos 120^{\circ }{)}^2)({\cos }^2(2\alpha )+{\sin }^2(2\alpha ))\\ & ={\left(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{3}\right)}^2+{\left(\frac{-1}{2}\right)}^2=\frac{3}{36}+\frac{1}{4}=\frac{1}{3}=(\tan 30^{\circ }{)}^2=(RHS{)}^2.\end{array}$$ $\square$