26th Turkish Mathematical Olympiad

Let $BC=a$, $u$ be the semiperimeter, the length of the altitude passing through $A$ be $h_a$, and the center and radius of $A$-excircle be $J_a$ and $r_a$, respectively. It is known that the points $A,D,J_a,E$ are collinear. We have $$AD=A{J}_a-J_{a}D=\sqrt{r_a^2+u^2}-r_a,DE=2r_a,AE=\sqrt{r_a^2+u^2}+r_a.$$ Therefore, we get $$\frac{AD}{AE}\cdot \frac{D{E}^2}{B{C}^2}=\frac{4r_a^2}{a^2}\cdot \frac{\sqrt{r_a^2+u^2}-r_a}{\sqrt{r_a^2+u^2}+r_a}={\left[\frac{2u}{a}\cdot \frac{r_a}{\sqrt{r_a^2+u^2}+r_a}\right]}^2.$$ Since the area of triangle $ABC$ is equal to $a{h}_a/2=r_a(u-a)$, we get $h_a/r_a=2(u-a)/a$. Let the feet of the perpendicular lines from $J_a$ and $A$ to the line $BC$ be $K$ and $L$, respectively. It is clear that $AL+J_{a}K\le A{J}_a$. This shows that $h_a+r_a\le \sqrt{r_a^2+u^2}$. Hence, we conclude that $$\frac{\sqrt{r_a^2+u^2}+r_a}{r_a}\ge \frac{h_a+2r_a}{r_a}=\frac{2u}{a}.$$ The last inequality completes the proof. The equality holds when the points $A,L,K,J_a$ are collinear, i.e., $AB=AC$.