Prove the excircle and right triangle inequality 
Let $\triangle{ABC}$ be a right triangle with $\angle{A}= 90^{\circ}$. Let $D$ be the intersection of the internal angle bisector of $\angle{A}$ with side $BC$ and $I_a$ be the center of the excircle of the triangle ABC opposite to the vertex $A$. Prove that $$\dfrac{DA}{DI_a} \leq \sqrt{2}-1.$$

I tried using reflections since $\angle{BI_aC} = 45^{\circ}$ we know that $I_a$ lies on a circle with $A$ as the center. I couldn't really find out how to incorporate the ratio. In case anyone wants a diagram to see this I drew one (ignore the scratch work to the side of it).

 A: Let circle O be the excircle (so O is its center). It must be tangent to its intersection with lines AB and AC by definition (call the intersection points X and Y). Then angles AXO and AYO are right angles, and by how X and Y were defined, so is XAY. Since XO and YO are radii of circle O, this means XO=YO, and thus quadridlateral XAYO is actually a square.
You also know that XO and YO are the same length as EO (where E is the point of tangency on BC), since EO is also a radius of circle O. Then you can show that equality holds in the problem statement when where AB=AC, and the ratio only gets smaller (or stays the same) if AB $\neq$ AC
A: Not the best solution but still:
Let E be point where excircle touches triangle $ABC$.
From right triangle $ EDI_a $ we have $ DI_a \geq r_a $
Using $ r_a=\frac{A}{s-a}$ and $A= \frac{1}{2}DA \cdot b \sin 45 + \frac{1}{2}DA \cdot c  \sin 45 = \frac{\sqrt2}{4} DA \cdot (b+c) $ where A is area of triangle $ABC$ we get 
$ DI_a \geq r_a = \frac{A}{s-a} $
$ \iff DA \leq DI_a \sqrt{2} \cdot \frac{b+c-a}{b+c} $
$ \iff DA \leq DI_a \sqrt2 (1-\frac{a}{b+c}) $
And since $ \frac{a}{b+c} = \frac{1}{\sin \alpha + \cos \alpha } \geq \frac{1}{\sqrt2}$ (Because $\sin \alpha + \cos \alpha \leq \sqrt2$) 
$ \iff DA \leq DI_a \sqrt2(1-\frac{a}{b+c}) \leq DI_a (\sqrt2-1)$ 
Q.E.D.
