# An Inequality with a Variety of Circumradii

### Solution 1

It's well known that $\displaystyle R_a=\frac{R}{2\cdot \cos A},\,$ and $\cos A\cdot \cos B\cdot \cos C\leq \frac{1}{8}.\,$ Thus,

\begin{align}\displaystyle &R_aR_bR_c=\frac{R^3}{8\cdot \cos A\cdot \cos B\cdot \cos C}\geq R^3,\\ &\frac{R_a^2}{R_b}+\frac{R_b^2}{R_c}+\frac{R_c^2}{R_a}\geq \frac{(R_a+R_b+R_c)^2}{R_a+R_b+R_c}=R_a+R_b+R_c,\\ &R_a+R_b+R_c\geq 3\cdot \sqrt[3]{R_aR_bR_c}=3\cdot \sqrt[3]{R^3},\,\text{by the AM - GM inequality}\\ &R_a+R_b+R_c\geq 3R \end{align}

as desired.

### Solution 2

By the property of central and inscribed angles subtended by the same arc, $\angle BOC =2A,\,$ thus, in the above diagram, $x=A.\,$ In $\Delta OBP,\,$ $OP=R \cos x,\,$ implying $OP=R \cos A.\,$ It follows that

$\displaystyle R_a=\frac{OB\cdot OC\cdot a}{4[\Delta OBC]}=\frac{aR^2}{4\cdot \frac{1}{2}a\cdot OP}=\frac{R^2}{2R\cos A}=\frac{R}{2\cos A}.$

Similarly, $\displaystyle R_b=\frac{R}{2\cos B}\,$ and $\displaystyle R_c=\frac{R}{2\cos C}.\,$ Therefore

\displaystyle \begin{align} LHS&=\frac{R}{2}\left(\frac{\cos B}{\cos^2 A}+\frac{\cos C}{\cos^2 B}+\frac{\cos A}{\cos^2 C}\right)\\ &\ge\frac{3R}{2}\sqrt[3]{\frac{1}{\cos A\cos B\cos C}},\,\text{due to the AM-GM inequality}\\ &\ge 3R,\,\text{because}\,\prod_{cycl}\cos A\leq \frac{1}{8} \end{align}

### Solution 3

It is well-known that

$\displaystyle R_a=\frac{R}{2\cos A}, R_b=\frac{R}{2\cos B}, R_c=\frac{R}{2\cos C}$

and

$\displaystyle \cos A\cdot \cos B\cdot \cos C\leq \frac{1}{8}.$

So by AM-GM inequality, we have

\displaystyle \begin{align}\frac{R_a^2}{R_b}+\frac{R_b^2}{R_c}+\frac{R_c^2}{R_a}&\geq 3\sqrt[3]{\frac{R_a^2R_b^2R_c^2}{R_bR_cR_a}}\\ &=3\sqrt[3]{R_aR_bR_c}=3\sqrt[3]{\frac{R^3}{8\cos A\cos B\cos C}}\\ &\geq 3\sqrt[3]{\frac{R^3}{1}}=3R. \end{align}

Equality holds when $\Delta ABC\,$ is equilateral.

### Solution 4

\displaystyle\begin{align} [\Delta BOC]&=\frac{1}{2}OA\cdot OB\cdot \sin (\widehat{BOC})=\frac{1}{2}R^2\sin 2A,\\ R_a&=\frac{OB\cdot OC\cdot BC}{4S[BOC]}=\frac{R\cdot R\cdot a}{4\cdot \frac{1}{2}R^2 \sin 2A}\\ &=\frac{a}{2\sin 2A}=\frac{2R\sin A}{2\cdot 2\sin A\cos A}=\frac{R}{2\cos A}. \end{align}

\displaystyle\begin{align} \frac{R_a^2}{R_b}+\frac{R_b^2}{R_c}+\frac{R_c^2}{R_a}&\overbrace{\geq}^{AM-GM}3\sqrt[3]{\frac{R_a^2}{R_b}\cdot \frac{R_b^2}{R_c}\cdot \frac{R_c^2}{R_a}}\\ &=3\sqrt[3]{R_aR_bR_c}=3\sqrt[3]{\frac{R^3}{8\cos A\cos B\cos C}}\\ &\geq 3\sqrt[3]{\frac{R^3}{8\cdot \frac{1}{8}}}=3R. \end{align}

### Acknowledgment

This is Dan Sitaru's problem from the Romanian Mathematical Magazine. Solution 1 by Mehmet Sahin (Turkey); Solution 2 is by Soumava Chakraborty (India); Solution 3 is by George Apostolopoulos (Greece); Solution 4 is by Dan Sitaru (Romania) who kindly provided a tex file with all the solutions.