Right Triangles and a Circle In a test sometime ago, I had to deal with the following question:

Let $\triangle ABC$ be a right angled triangle.  Let $\triangle DEF$ be a right triangle inscirbed in the incircle of $\triangle ABC$. Find the smallest possible ratio of $\frac{[ABC]}{[DEF]}$ where $[ABC]$ denotes the area of $\triangle ABC$.

What I did was to break the problem into 2 parts:
$1$. Find the largest right triangle with a circumcircle of a fixed radius.
$2$. Find the smallest right triangle with a incircle with a fixed radius.
For the first part, I was able to prove that the isosceles right triangle had the largest area using similar triangles but I got stuck on the second part during the test and was unable to complete the question. I have a suspicion that it is also an isosceles right triangle but I could not prove it.
I have looked up some info on the incircle and tried the following method:
Let $P_{ABC}$ denote the perimeter of $\triangle ABC$ and let $\angle BAC = 90^{\circ}$.
WLOG Let the inradius $r$ of $\triangle ABC$ be 1
Since $[ABC] = \frac{1}{2} \cdot P_{ABC} \cdot r$ and $[ABC] = \frac{1}{2} \cdot AB \cdot AC$, we have:
$$
\begin{align*}
P_{ABC} & = AB \cdot AC\\
(AB + AC + BC) & = AB \cdot AC\\
\end{align*}
$$
This is where I got stuck. Is it possible to further develop this method or is there a simpler method all together?

If a right triangle $DEF$ is inscribed in a circle with radius $r$, its maximum area is given by $r^2$, since the longest side of such a triangle is also a diameter of the circumcircle of $DEF$. It follows that $\frac{[ABC]}{[DEF]}$ is lower bounded by $\frac{[ABC]}{r^2}$ where $r$ is the inradius of $ABC$. On the other hand, if the lengths of the legs of $ABC$ are $x$ and $y$, we have $[ABC]=\frac{xy}{2}$ and $r=\frac{2[ABC]}{x+y+\sqrt{x^2+y^2}}=\frac{xy}{x+y+\sqrt{x^2+y^2}}$, so:
$$ \frac{[ABC]}{[DEF]}\geq \frac{\left(x+y+\sqrt{x^2+y^2}\right)^2}{2xy}\geq\frac{\left(2\sqrt{xy}+\sqrt{2}\sqrt{xy}\right)^2}{2xy}=\color{red}{3+2\sqrt{2}} $$
by the QM-AM-GM inequality:
$$ \sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}\geq \sqrt{xy}. $$
Equality is achieved at $x=y$ as expected.
