<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] Recall that $$1 = r = \frac{abc}{4sR} = \frac{abc}{4 \cdot \frac{a+b+c}2\cdot \frac{c}2} \implies ab = a + b+ c$$
where $R = \frac{c}2$ is the circumradius.
Now we have
$$ab = a + b + c = a + b + \sqrt{a^2 + b^2} \ge 2\sqrt{ab} + \sqrt{2ab} = (2+\sqrt{2})\sqrt{ab}$$
so $ab \ge (2+\sqrt{2})^2 = 2(3+2\sqrt{2})$.
The equality is attained if and only if $a = b = 2+\sqrt{2}$ so the minimum area is indeed $$\frac{ab}2 = 3+2\sqrt{2}$$
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