An inequality satisfied by the circumradius and inradius of a right triangle

Question

This should be trivial, but I am unable to show that $R \geq (1+\sqrt{2})r$ for a right triangle. Where $R$ is the circumradius and $r$ is the inradius of a right triangle.

Answer 1

Let $a,b,c$ be the sides of the triangle ($c$ is the hypotenuse). Then the circumradius is $c/2$, while The inradius has the formula $$r = \frac{2\Delta}{P}$$ where $\Delta$ is the area and $P$ is the perimeter. So, we have $$r = \frac{ab}{a+b+c}$$ for a right triangle. Now, we want to show that $(1+\sqrt{2})\frac{ab}{a+b+c} \leq (c/2)$, or $$ (2+2\sqrt{2}) ab \leq c(a+b+c) \hspace{2in} (1) $$

Now, by AM-GM, $c=\sqrt{a^2+b^2} \geq \sqrt{2ab}$, and $a+b \geq 2 \sqrt{ab}$. Plugging in these inequalities, you can verify $(1)$.

Answer 2

First, note that for any right triangle with legs $a$ and $b$ and hypotenuse $c$, $r=\frac{1}{2}(a+b-c)$.

You should be able to convince yourself that, for a given hypotenuse length $c$ (and hence a given circumradius, since the circumcenter of any right triangle is the midpoint of the hypotenuse), the inradius is maximized when the right triangle is isosceles, so $a=b$. For convenience (all such triangles are similar), let $c=2$ so that $a=b=\sqrt{2}$ and $R=1$. Now, $r=\frac{1}{2}(a+b-c)=\frac{1}{2}(2\sqrt{2}-2)=\sqrt{2}-1$, so $(1+\sqrt{2})r=(1+\sqrt{2})(\sqrt{2}-1)=1=R$. That is, in the case where $r$ is maximal, equality holds for your inequality. So, for all right triangles, $R\ge(1+\sqrt{2})r$.