similar problem iran round 2 p4.

First off we will prove a useful lemma. We shall denote by $a,b,c$ the sides of the triangle as $BC,CA,AB$ respectfully. It is also obvious that $AD$ is the angle bisector. LEMMA: The length $d=AD$ in any given triangle is $d=\frac{bc}{(b+c)^2}((b+c)^2-a^2)$ PROOF: It follows from the angle bisector theorem that $\frac{BD}{DC} = \frac{c}{b}$, from here we do the following: $$ \frac{BD}{DC}=\frac{c}{b}$$$$\frac{BD}{DC}+1=\frac{c}{b}+1$$$$\frac{a}{DC}=\frac{b+c}{b}$$$$DC= \frac{ab}{b+c}$$Similarly we get that $BD=\frac{ac}{b+c}$ Now applying Stewart's theorem we get that: $$b^2.BD+c^2.DC=d^2.a+BD.BC.a$$Now plugging in what we got, and we finally get the desired result: $$d^2=\frac{bc}{(b+c)^2}((b+c)^2-a^2)$$In our case we are dealing here with a right-angled triangle so that means we can simplify this expression a little bit: $$d^2=\frac{bc}{(b+c)^2}(b^2+c^2+2bc-a^2)=\frac{bc}{(b+c)^2}.2bc \implies d=\frac{bc}{b+c}\sqrt 2$$ So now back to the problem. First notice that the triangle $\triangle I_aEA$ is a right-isoceles one, so that means that $AI_a = IE\sqrt 2 = \frac{s}{s-a}.r \sqrt 2 = s\sqrt 2$ Also notice that the points $A,D,I_a$ are colinear. So that means that $AD+DI_a=AI_a$ From here we have that $DI_a=AI_a-AD=s\sqrt 2 - \frac{bc}{b+c} \sqrt 2 = (s-\frac{bc}{b+c})\sqrt 2$ So now we have that the ratio $\frac{AD}{DI_a}$ can be expressed as $\frac{\frac{bc}{b+c}}{s-\frac{bc}{b+c}} \leqslant \sqrt 2 -1$ So with some algebra we gotta prove that the following holds: $$\frac{(a+b+c)(\sqrt 2 -1)}{2} \geqslant \frac{bc}{b+c}\sqrt 2$$Now again with some algebra we can reduce this to: $$(ab+ac+b^2+c^2)(\sqrt 2 -1) \geqslant 2bc$$But remember that we have $b^2+c^2=a^2$ so that means we have the following: $$a(b+c+a)(\sqrt 2 -1) \geqslant 2bc$$Which is reduced to the following: $$as(\sqrt 2-1) \geqslant bc$$So now we use the $AM-GM$ inequalits on $a$ ans on $s$ and we get that $a \geqslant \sqrt{2bc}$ and that $s \geqslant \frac{\sqrt{bc}(2+\sqrt 2)}{2}$,So from here we have that: $$LHS \geqslant bc\frac{2+\sqrt 2}{2} \sqrt 2 (\sqrt 2 -1) = bc \frac{2(\sqrt 2 +1)(\sqrt 2 -1)}{2} = bc $$Thus the inequality holds . . .

Much simpler: Note that we need to show that $\frac{AI_A}{DI_A} \le \sqrt{2}$. However $DI_A \ge \mathrm{dist}(I_A, BC) = \mathrm{dist}(I_A, CA) = \frac{AI_A}{\sqrt{2}}$, whence we are done.