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
$$AD = AJ_a - J_a D = \sqrt{r_a^2 + u^2} - r_a, \quad DE = 2r_a, \quad AE = \sqrt{r_a^2 + u^2} + r_a.$$

Therefore, we get
$$\frac{AD}{AE} \cdot \frac{DE^2}{BC^2} = \frac{4r_a^2}{a^2} \cdot \frac{\sqrt{r_a^2 + u^2} - r_a}{\sqrt{r_a^2 + u^2} + r_a} = \left[ \frac{2u}{a} \cdot \frac{r_a}{\sqrt{r_a^2 + u^2} + r_a} \right]^2.$$
Since the area of triangle $ABC$ is equal to $ah_a/2 = r_a(u-a)$, 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 $AL + J_a K \le AJ_a$. 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., $AB = AC$.