Selection and Training Session

We prove the statement for $\angle C\ge 36^{\circ }$. Let $\angle ABC=\angle BAC=2x$. We have $DC+DL\le BC\iff \frac{BC}{DC}\ge 1+\frac{DL}{DC}$. Now, by the law of sines, $$\begin{gathered} \frac{BC}{DC}=\frac{\sin \angle BDC}{\sin \angle DBC}=\frac{\sin 2x}{\sin x}=2\cos x, \\ \frac{DL}{DC}=\frac{\sin 3x}{\sin x}=\frac{1}{3-4{\sin }^{2}x}=\frac{1}{4{\cos }^{2}x-1}, \end{gathered}$$ so we need to prove that $2\cos x\ge 1+\frac{1}{4{\cos }^{2}x-1}=\frac{4{\cos }^{2}x}{4{\cos }^{2}x-1}$. Note that $x\le 36^{\circ }$ since $\angle C\ge 36^{\circ }$, so $\cos x>\frac{1}{2}$ and $4{\cos }^{2}x-1>0$. Therefore the inequality is equivalent to $$4{\cos }^{2}x-1\ge 2\cos x\iff 4{\cos }^{2}x-2\cos x-1\ge 0\iff \cos x\ge \frac{-1+\sqrt{5}}{4},$$ which is true for $x\le 36^{\circ }$.