# Euler Inequality in Triangle

If $R$ and $r$ are the circumradius and, respectively, the inradius of a triangle, then $R\ge 2r.$

This is known as (another) Euler's inequality. The inequality is an immediate consequence of Euler's identity, $OI^2 = R^2 - 2Rr,$ where $O$ is the circumcenter and $I$ the incenter of the triangle.

Here's another proof.

### Proof

(Elias Lampakis, Am Math Monthly, 122 (9), November 2015, p 892.)

Let $a,b,c$ denote the line lengths, $s$ the semiperimeter, $r_a,\;$ $r_b,\;$ $r_c,\;$ the exradii, $[ABC]\;$ the area of $\Delta ABC,$

$\displaystyle [ABC]=\sqrt{s(s-a))(s-b)(s-c)}=sr=\frac{abc}{4R}=r_x(s-x),$ $x\in\{a,b,c\}.$

Then

\displaystyle\begin{align} 4rr_a&=4\frac{[ABC]^2}{s(s-a)}=4(s-b)(s-c)\\ &=(a+c-b)(a+b-c)=a^2-(b-c)^2\\ &\le a^2. \end{align}

Similarly, $4rr_b\le b^2$ and $4rr_c\le c^2.$ Multiplying all three we see that

\begin{align} 64r^3r_ar_br_c &\le a^2b^2c^2 \Leftrightarrow\\ 64r^3[ABC]^3 &\le 16R^2[ABC]^2(s-a)(s-b)(s-c) \Leftrightarrow\\ 4r^4s &\le R^2(s-a)(s-b)(s-c) \Leftrightarrow\\ 4r^4s^2 &\le R^2[ABC]^2=R^2s^2r^2 \Leftrightarrow\\ 4r^2 &\le R^2, \end{align}

and Euler's inequality follows.