Minimal area of a right triangle with inradius $1$

Question

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.

Answer 1

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 $$\frac{ab}2 = 3+2\sqrt{2}$$

Answer 2

$$1=r=\frac{a+b-c}{2}=\frac{a+b-\sqrt{a^2+b^2}}{2}.$$

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 $$\min_{r=1}S=3+2\sqrt2.$$