We consider triangles with perimeter 1 and one of angles $120^o$. Find the length of the sides of this triangle for which the area of the inscribed circle is the largest.
Let $\alpha = 2\pi/3$ and use the fact that $r=A/s=2A=2\cdot 1/2 \cdot b \cdot c \cdot \sin{\alpha}=\frac{\sqrt{3}}{2}bc$. So we have to find how to maximise $bc$ while respecting the condition $a+b+c=1$ and the fact that it is a triangle. I don't know how to continue from here though.
Let $a$ be the side opposite to the $120º$ angle. Then, by the cosine law and knowing that $a = 1-(b+c)$, $$a^2=b^2+c^2-2bc\cos (\frac{2\pi}{3})=b^2+c^2+bc$$ $$\require{cancel}1-2(b+c)+\cancel{b^2}+2bc+\cancel{c^2}=\cancel{b^2}+\cancel{c^2}+bc$$ $$bc=2b+2c-1$$ $$b=\frac{2c-1}{c-2}$$
So we must find the maxima of $bc=\frac{c(2c-1)}{c-2}$, which are at $c=2\pm\sqrt{3}$, but only $c=2-\sqrt{3}<1$. Then, as expected because of the symmetry of the problem, $b=c=2-\sqrt{3}$ and $a=2\sqrt{3}-3$.