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
$$BD \le CE \Leftrightarrow AB \tan \alpha \le BC \tan 30^\circ \Leftrightarrow \frac{AB}{BC} \tan \alpha \le \tan 30^\circ$$
$$\Leftrightarrow \tan \alpha \frac{\sin(120^\circ - 2\alpha)}{\sin(2\alpha)} \le \tan 30^\circ \quad (\text{Law of Sines})$$
$$\Leftrightarrow \frac{\sin \alpha}{\cos \alpha} \frac{\sin(120^\circ - 2\alpha)}{2 \sin \alpha \cos \alpha} \le \tan 30^\circ$$
$$\Leftrightarrow \sin(120^\circ - 2\alpha) \le 2\cos^2 \alpha \cdot \tan 30^\circ$$

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