Triangle $ABC$ has $\angle C=90^\circ$ and $BC=4$. Point $D$ is the midpoint of $\overline{BC}$. What is the largest possible value of $\tan\angle BAD$?
Place $C=(0,0), B=(4,0), A=(0,t)$, $D=(2,0)$. Then $\tan\angle BAD=\dfrac{2t}{t^2+8}$ (via the cross/dot product of $\vec{AB},\vec{AD}$).
$\frac{2t}{t^2+8}=\frac2{t+\frac8t}$, and by AM-GM $t+\frac8t\ge4\sqrt2$ with equality at $t=2\sqrt2$. So the maximum value is $\dfrac{2(2\sqrt2)}{8+8}=\dfrac{4\sqrt2}{16}=\boxed{\dfrac{\sqrt2}4}$.
