<s>[INST] Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.

My attempt:

 [/INST] We'll prove that for all triangle 
$$r\leq\sqrt{\frac{A}{3\sqrt3}}.$$
Indeed, let $AB=c$, $AC=b$ and $BC=a$. 
Hence, $A=\frac{1}{2}ra+\frac{1}{2}rb+\frac{1}{2}rc$, which gives $r=\frac{2A}{a+b+c}$.
Thus, we need to prove that
$$\frac{2A}{a+b+c}\leq\sqrt{\frac{A}{3\sqrt3}}$$ or
$$12\sqrt3A\leq(a+b+c)^2.$$
But by Heron formula we have
$$A=\sqrt{p(p-a)(p-b)(p-c)},$$
where $p=\frac{a+b+c}{2}$.
Thus, we need to prove that
$$(a+b+c)^2\geq12\sqrt3\sqrt{\frac{a+b+c}{2}\cdot\frac{a+b-c}{2}\cdot\frac{a+c-b}{2}\cdot\frac{b+c-a}{2}}$$ or
$$(a+b+c)^3\geq27(a+b-c)(a+c-b)(b+c-a)$$ or
$$\frac{(a+b-c)+(a+c-b)+(b+c-a)}{3}\geq\sqrt[3]{(a+b-c)(a+c-b)(b+c-a)},$$
which is AM-GM.
The equality occurs for $$a+b-c=a+c-b=b+c-a$$ or
$$a=b=c,$$
which says that $r$ gets a maximal value, when our triangle is an equilateral triangle,
which says that $\measuredangle BAC=60^{\circ}$.
Done! 

If the following is obvious for you, then I am ready to delete it.

The equality occurring in our AM-GM just for $a+b-c=a+c-b=b+c-a$ we can understand by the following way.
Let $a+b-c=x^3$, $a+c-b=y^3$ and $b+c-a=z^3$.
Hence, $x$, $y$ and $z$ are positives and the equality case gives
$$\frac{x^3+y^3+z^3}{3}=xyz$$ or
$$x^3+y^3+z^3-3xyz=0$$ or
$$x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0$$ or
$$(x+y)^3+z^3-3xy(x+y+z)=0$$ or
$$(x+y+z)((x+y)^2-(x+y)z+z^2)-3xy(x+y+z)$$ or
$$(x+y)^2-(x+y)z+z^2-3xy=0$$ or
$$x^2+y^2+z^2-xy-xz-yz=0$$ or
$$2x^2+2y^2+2z^2-2xy-2xz-2yz=0$$ or
$$(x-y)^2+(x-z)^2+(y-z)^2=0$$ or
$$x=y=z,$$
which gives $a=b=c$.
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