<s>[INST] Minimal area of a right triangle with inradius $1$ I got this question, solved it, then forgot how I solved it.

What is the minimal area of a right triangle with inradius $1$?

My attempt:
$r=\frac{a+b-c}2$, so $a+b=c+2$
$a^2+b^2=c^2$
This gives $ab=2(c+1)$
I remember using the AM-GM to prove that equality held, thus giving $a=b=2+\sqrt2$ and getting the area as $3+2\sqrt2$, but I can't do it now. Please help.
 [/INST] $$1=r=\frac{a+b-c}{2}=\frac{a+b-\sqrt{a^2+b^2}}{2}.$$
We'll prove that
$$S\geq3+2\sqrt2.$$
Indeed, we need to prove that:
$$\frac{ab}{2}\geq(3+2\sqrt2)\left(\frac{a+b-\sqrt{a^2+b^2}}{2}\right)^2$$ or
$$(a+b+\sqrt{a^2+b^2})^2\geq2(3+2\sqrt2)ab,$$
which is true by AM-GM:
$$(a+b+\sqrt{a^2+b^2})^2\geq(2\sqrt{ab}+\sqrt{2ab})^2=2(3+2\sqrt2)ab.$$
The equality occurs for $a=b=2+\sqrt2,$ which says that
$$\min_{r=1}S=3+2\sqrt2.$$
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