[geogebra]dfd79e0a192fb0b91554946f653cbd6dfea9e05b[/geogebra] $F$ is the point of the intersection of $(CE)$ and $(BD)$ since $\angle ABC = 60$ and $(FB)\perp(AB)$ then $\angle FBE = \angle FBC+\angle CBE = 60$ $...(1)$ then we have $(BC)\perp(FC)$, which means: $\angle BFC = 180 - (\angle FBC + \angle FCB) = 60$ $...(2)$ from $(1)$ and $(2)$ we get that $\triangle FBE$ is equilateral. By putting $G$ the midpoint of the side $[BF]$ we get: $(x= \angle GDE)$ $cos(x)=\frac{GD}{DE}$ and since $D\in[BF]$ then $DE$ $\ge$ $GD$ that means: $0 \le \frac{GD}{DE} \le 1$ $\Rightarrow 0 \le cos(x) \le 1$ $\Rightarrow x \le 90$ $\Rightarrow \angle GDE \le 90$ $\Rightarrow \angle BDE \ge 90$ and since $\angle DBE = 60$ then $\angle BED \le 30$

It can be done with AM-GM easily.

Let $M$ be the intersection of $BD$ and $CE$, then it is easy to see that $ \triangle MBE $ is equilateral. Now let N be the foot of the perpendicular from the point $D$ to $AC$. In $\triangle DNC$, $ \angle DNC = 90$ so $DN\le DC$, equality holds then $N=C$. But $DN=BD$ since $AD$ is the bisector of $ \angle BAC $. It means that $DC\le DB$ and since point $C$ is the middle of $ME$ and $ \triangle MBE $ is equilateral, $DB\le \frac {MB} {2}$.Which means that $ \angle BED\le 30 $.

Let $\angle A= 2 \angle \alpha$ then $\left\{\begin{matrix} & \\ BE = \frac{2} {\sqrt{3}} . BC & BD = AB . \tan \: \alpha \end{matrix}\right.$ but $tan \: \alpha = \frac{\sin A}{\cos A +1}$ so $BD \leq \frac{BE}{2} $ reduce to $\sqrt{3} \sin \: C = \sqrt{3} sin \: (120 -A) \leq \cos \: A +1$ which is equal to : $ \sin \: (A+30) \leq 1$

Let S be reflection of E across BC. we have ∠ABE = ∠EBC = ∠CBD = 30 so S lies on BD and ∠SBE = 60 so triangle BES is regular. Let K be midpoint of segment BS. instead we can prove BD ≤ BK. D lies on angle bisector of ∠A so BD equals distance of D from AC and it's less or equal with CD. so now we have BD ≤ CD so ∠DBC = 30 ≥ ∠DCB. we have ∠KCB = 30 and ∠DCB ≤ 30 so BD ≤ BK. we're Done.