I did a little work but I couldn't get a solution: First I split up the smaller right triangle into two isosceles triangles, both with the equal side as r. Then I used sin formula and added and simplified to get S' = 2sin(x)cos(x)r^2. I also got S = r^2. (r is the radius of the incircle and x is the non-ninety degree angle of the second triangle). I tried to use AM-GM inequality on the S" part of the ratio but it didn't work out. Got stuck after that.

Given $A(0,0),B(b,0),C(0,c)$, area of $\triangle ABC\ :\ S=\frac{1}{2}bc$. Angle bisector of $A\ :\ y=x$. Angle bisector of $B\ :\ cx+(b+\sqrt{b^{2}+c^{2}})y-bc=0$. Center of the inscribed circle: $O(\frac{bc}{b+c+\sqrt{b^{2}+c^{2}}},\frac{bc}{b+c+\sqrt{b^{2}+c^{2}}})$. His circle is circumscriped to a triangle with hypothenusa $2 \cdot \frac{bc}{b+c+\sqrt{b^{2}+c^{2}}}$ and height $\frac{bc}{b+c+\sqrt{b^{2}+c^{2}}}$. Area $S'=\frac{1}{2} \cdot 2(\frac{bc}{b+c+\sqrt{b^{2}+c^{2}}})^{2}$. Then $\frac{S}{S'}=\frac{bc}{2(\frac{bc}{b+c+\sqrt{b^{2}+c^{2}}})^{2}}=\frac{b^{2}+c^{2}+bc+(b+c)\sqrt{b^{2}+c^{2}}}{bc} \geq 3+2\sqrt{2}$. Equality if $b=c$.