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

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.

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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 $(1)$.

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@J.M. corrected, thanks! – Srivatsan Jul 30 '11 at 4:22
Okay, I gave a +1 for the algebraic approach. :) – J. M. Jul 30 '11 at 4:24

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, $R\ge(1+\sqrt{2})r$.

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Can you show why the inradius is maximized when the triangle is isosceles? I know intuitively it is true, but how would one prove it? – picakhu Jul 30 '11 at 4:12
@picakhu: Fix $c$ and let $a$ vary in the interval $(0,c)$. From the Pythagorean Theorem, $b=\sqrt{c^2-a^2}$; $r=\frac{1}{2}(a+b-c)$ so as a function of $a$, $r(a)=\frac{1}{2}(a+\sqrt{c^2-a^2}-c)$. Take the derivative and look for sign changes. The only sign change in the derivative occurs when $a=\frac{c}{\sqrt{2}}$, which makes the triangle isosceles. – Isaac Jul 30 '11 at 4:21
@picakhu: Less formally, for a fixed $c$, as $a\to0$ or alternately $b\to0$, $r\to0$. $r$ is positive between those extremes, so there is a maximum. The problem is symmetric with swapping $a$ and $b$, so the maximum should be exactly between those extremes, so when $a=b$. – Isaac Jul 30 '11 at 4:23
...and I upvoted this for the geometric way of looking at the problem. – J. M. Jul 30 '11 at 4:25