Let $B', C'$ be points on lines $AB$ and $AC$ respectively, such that $B'C' \parallel BC$ and $B'C'$ is also tangent to $A$ excircle. It is clear that there is a homothethy centered at A sending $\triangle ABC$ to $ \triangle AB'C'$ and from the fact that $\frac{r_a}{r}=\frac{s}{s-a}$ we have that this homothethy has a coefficient of $\frac{s}{s-a}$ (here $a=\overline{BC}$ and $s$ semiperimeter). Let the intersection of the line $AB$ and $A$-excircle be $ X$. It is a well-known fact that $AX=s$. Now, by the power of point $A$ w.r.t to $A$-excircle we get that $AD \cdot AE=AX^2 \Longleftrightarrow \frac{AD}{AE}= \frac{AX^2}{AE^2}$. Lemma: In every triangle $\triangle ABC$ let $I$ be the incenter and let the angle bisector at $A$ intersect the incircle at 2 points $P, Q$ such that $AQ>AP$ then $AQ>h$ where $h$ is the altitude erected from $ A$. Proof of lemma: Let $Y$ be the foot of $I$ into $BC $, we have that $IY=IQ$ and by triangle inequality $AQ=AI+IQ=AI+IY \geq AY \geq h$. Now by the above lemma in triangle $AB'C' $ we have that $AE\geq h$ (here $h$ is now altitude from $A$ to $B'C'$ ). The area of $\triangle AB'C'$ is $$r_a \cdot \frac{s^2}{s-a}= \frac{A'B' \cdot h }{2}= \frac {AB \cdot s \cdot h}{2(s-a)} \leq \frac{AB \cdot AE \cdot s}{2(s-a)}$$This gives that $2r_a \cdot s \leq BC \cdot AE$ and since $2r_a=DE$ and $s=AX$ this is equivalent to $\frac{AX}{AE} \leq \frac{BC}{DE}$ and we have that $$\boxed{\frac{AD}{AE}=\frac{AX^2}{AE^2} \leq \frac{BC^2}{DE^2}}$$$\blacksquare$ Equality at $AE=h$ (i.e when $AB=AC$)

Claim: If $|AB|=|AC|$, then $\frac{|AD|}{|AE|}=\frac{|BC|^2}{|DE|^2}$. Proof: Let $I_A$ be the excenter wrt $A$. Let $|AB|=|AC|=x$ and $\angle BAD=\angle CAD=2\alpha$. By angle chasing, one can find that $\frac{|AD|}{|AE|}=\frac{1-\sin{(2\alpha)}}{1+\sin{(2\alpha)}}$ and $\frac{|BC|^2}{|DE|^2}=\frac{|BD|^2}{|DI_A|^2}=\frac{\sin^2{(45^\circ-\alpha)}}{\sin^2{(45^\circ+\alpha)}}$. You can easily verify that $\frac{1-\sin{(2\alpha)}}{1+\sin{(2\alpha)}}=\frac{\sin^2{(45^\circ-\alpha)}}{\sin^2{(45^\circ+\alpha)}}$. Hence, $\frac{|AD|}{|AE|}=\frac{|BC|^2}{|DE|^2}$, as desired. $\;\;\;\;\square$ Now, WLOG $|AC|>|AB|$. Let $I_A$ be the excenter wrt $A$ and assume that the tangent at $D$ to the excenter intersects $AB$ and $AC$ at $X$ and $Y$, respectively. We know that $|AX|=|AY|$ and excircle of $ABC$ of wrt $A$ is also the excircle of $AXY$ wrt $A$. Now, if we apply our claim to $AXY$, we can find that $\frac{|AD|}{|AE|}=\frac{|XY|^2}{|DE|^2}$. Hence, it suffices to prove that $\frac{|XY|^2}{|DE|^2}<\frac{|BC|^2}{|DE|^2}\Leftrightarrow |BC|>|XY|$. Check that the triangles $ABC$ and $AXY$ have the same perimeter. Lemma: Given a triangle $ABC$. We have $\frac{|BC|}{\sin{(\frac A2)}}\ge |AB|+|AC|$ and equality occurs iff $|AB|=|AC|$. Proof: Take a point $C'$ on $AB$ such that $|AC'|=|AC|$ and $A\in [BC']$. We have $\angle AC'C=\frac A2$. Hence, $$\frac{|BC|}{\sin{(\frac A2)}}=\frac{|BC'|}{\sin{(BCC')}}=\frac{|AB|+|AC|}{\sin{(BCC')}}\ge |AB|+|AC|$$Equality occurs iff $\angle BCC'=90^\circ\Leftrightarrow |AB|=|AC|$, done. Thus, we know that $\frac{|BC|}{\sin{(\frac A2)}}>|AB|+|AC|\Leftrightarrow |BC|+\frac{|BC|}{\sin{(\frac A2)}}>|AB|+|AC|+|BC|$. By the Law of Sines, one can find that the perimeter of $AXY$ equals to $|XY|+\frac{|XY|}{\sin{(\frac A2)}}$. Since the triangles $ABC$ and $AXY$ have the same perimeter, we find that $$|BC|+\frac{|BC|}{\sin{(\frac A2)}}>|XY|+\frac{|XY|}{\sin{(\frac A2)}}\Rightarrow |BC|>|XY|\text{, as desired}.$$ Equality occurs iff $|AB|=|AC|$.

Let $(I_A)$ be tangent to $AC$ at $T$. Line passing through $D$ cuts $AB,AC$ at $K,L$. Then $DLI_A \sim DTZ \implies $ $$\frac{AD}{AE}=\frac{DT^2}{TZ^2}=\frac{KL^2}{DE^2}\le \frac{BC^2}{DE^2}$$

Let the tangency point of $A$ excircle and $AB$ be $X$, then $$\frac{AD}{AE}=\frac{AX^2}{AE^2}$$so we need to prove that $$\frac{s}{AE} \leq \frac{BC}{2r_a}$$where $s $ is the semiperimeter of $\triangle ABC$. $$s\cdot 2r_a = 4[AXI_A]$$and $$BC \cdot AE \geq BC\cdot (AH +2r_a) =2[ABC]+2[I_ABC]=4[AXI_A]$$where $H$ is projecton of $A$ onto $BC$. This finishes the proof.

Let the center of the excircle be $ I $ and the feet of the perpendiculars from $ I $ (or tangency points) to $ AB $ and $ AC $ be $ X $ and $ Y $ respectively. And let $ \angle BAC = 2a, \angle ABC = 2b, \angle ACB = 2c $ and $|DI|=r$. (Note that $a+b+c=90^\circ$, and $\angle BIX = b$ and $\angle CIY = c$ .) Finally, we can start the proof. We need to prove: $\frac{|BC|^2}{|DE|^2} \ge \frac{|AD|}{|AE|}= \frac{|AD||AE|}{|AE|^2}= \frac{|AX|^2}{|AE|^2}$ which is obviously equivalent to $\frac{|AX|}{|AE|} \le \frac{|BC|}{|DE|}$. And: $\frac{|AX|}{|AE|} \le \frac{|BC|}{|DE|}= \frac{|BX|+|CY|}{|DE|} =\frac{r\tan(b)+r\tan(c)}{2r}=\frac{\tan(b)+\tan(c)}{2}=\frac{\sin(b+c)}{2\cos(b)\cos(c)}$ Also $|AX|=|AI|\cos(a)=|AI|\sin(b+c)$ So we should prove: $\frac{|AI|\sin(b+c)}{|AE|} \le \frac{\sin(b+c)}{2\cos(b)\cos(c)}$ $\Leftrightarrow 2\cos(b)\cos(c) \le \frac{|AE|}{|AI|}=\frac{|AI|+|IE|}{|AI|}=1+ \frac{r}{|AI|}=1+ \sin(a)= 1+ \cos(b+c)$ $\Leftrightarrow 1 \ge 2\cos(b)\cos(c)-\cos(b+c)=\cos(b-c)$, which is always true, so we are done. Equality occurs if and only if we have $b-c=0$, meaning $|AB|=|AC|$ .

Let the $A$-excircle touch $AB$ at $X$ and let $|I_AX| = r_a$, where $I_A$ is the $A$-excenter. Note that since the line $AE$ bisects $\angle BAC$, $I_A$ lies on $AE$. Also, we have $|AX| = s$, where $s$ is the semiperimeter of $\triangle ABC$. Lastly, call $|BC| = a$. Now, $|AD| = \sqrt{s^2 + r_a^2} - r_a$, $|AE| = \sqrt{s^2 + r_a^2} + r_a$, $|BC| = a$ and $|DE| = 2r_a$. The requested inequality reduces then to showing that $$\dfrac{\sqrt{s^2 + r_a^2} - r_a}{\sqrt{s^2 + r_a^2} + r_a} \leq \dfrac{a^2}{4r_a^2} \Longleftrightarrow (4r_a^2-a^2)^2(s^2+r_a^2) \leq r_a^2(4r_a^2+a^2)^2 \Longleftrightarrow 4r_a^2(s-a) \leq a^2s$$which since $(s-a)r_a = sr$ ($r$ is the inradius of $\triangle ABC$) simplifies to $$ 4r_ar \leq a^2 $$Finally, as $r_ar = \dfrac{S}{s-a} \cdot \dfrac{S}{s} = (s-b)(s-c)$ ($S$ is the area of $\triangle ABC$), $$ 4(s-b)(s-c) \leq a^2 \Longleftrightarrow (2s-b-c)^2-4(s-b)(s-c) \geq 0 \Longleftrightarrow (c-b)^2 \geq 0 $$This finishes the proof.

This problem is a bit of a ruse: it presents itself as a triangle problem but isn't really about triangles. Sketch: Fix the angle $A$ and the excircle $\omega_A$ (with center $I_A$ and radius $r_A$), and allow $B$ and $C$ to vary such that segment $\overline{BC}$ is tangent to $\omega_A$. Let $D$ be the tangency point. Then $\angle BI_AD$ and $\angle CI_AD$ are two acute angles which sum to the constant angle $\angle B_IAC = \tfrac{90^\circ + \angle A}2$, so by Jensen's Inequality $BC = r_A(\tan \angle BI_AD + \angle CI_AD)$ is minimized when both angles are equal to each other. This occurs when triangle $BAC$ is isosceles. It suffices to show that $BC^2 = DE^2 \cdot\tfrac{AD}{AE}$ in the case where triangle $ABC$ is isosceles, which BarisKoyuncu does in post #4.

Great solutions