An Inequality with Inradius and Circumradii

Problem

Inradius - one, curcumradii - many

Solution

In every triangle with the side lengths $a,b,c,\,$ area $S,\,$ and circumradius $R,\,$ we have $abc=4RS.\,$ So, in $\Delta IAB,\,$

$\displaystyle R_c=\frac{AI\cdot BI\cdot AB}{4[\Delta IAB]} = \frac{\displaystyle\frac{r}{\sin\frac{A}{2}}\cdot\frac{r}{\sin\frac{B}{2}}\cdot a}{4\cdot\displaystyle\frac{ar}{2}} = \frac{r}{\displaystyle 2\sin\frac{A}{2}\sin\frac{B}{2}},$

Therefore, $\displaystyle R_c=\frac{r}{\displaystyle 2\sin\frac{A}{2}\sin\frac{B}{2}}\,$ and, similarly, $\displaystyle R_b=\frac{r}{\displaystyle 2\sin\frac{C}{2}\sin\frac{A}{2}}\,$ and $\displaystyle R_a=\frac{r}{\displaystyle 2\sin\frac{B}{2}\sin\frac{C}{2}}.\,$ Thus, using Bergstrom's inequality,

$\displaystyle\begin{align} \sum_{cycl}R_a &=\frac{r}{2}\sum_{cycl}\frac{1}{\displaystyle \sin\frac{B}{2}\sin\frac{C}{2}} \ge\frac{r}{2}\cdot\frac{(1+1+1)^2}{\displaystyle \sum_{cycl}\sin\frac{B}{2}\sin\frac{C}{2}}\\ &\ge\frac{r}{2}\cdot\frac{9}{\displaystyle \frac{1}{3}\left(\sum_{cycl}\sin\frac{A}{2}\right)^2} =\frac{r}{2}\cdot\frac{27}{\displaystyle \left(\frac{3}{2}\right)^2}=\frac{r}{2}\cdot\frac{4}{9}\cdot 27\\ &=6r. \end{align}$

It follows that

(1)

$\displaystyle \sum_{cycl}R_a=6r.$

Further,

$\displaystyle \begin{align} \sum_{cycl}\frac{R_a}{R_bR_c} &= \sum_{cycl}\frac{\displaystyle \frac{r}{2\sin\frac{B}{2}\sin\frac{C}{2}}}{\displaystyle \sin^2\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=\frac{2}{r}\sum_{cycl}\sin^2\frac{A}{2}\\ &=\frac{1}{r}\sum_{cycl}(1-\cos A)=\frac{3}{r}-\frac{1}{r}\sum_{cycl}\cos A\\ &=\frac{3}{r}-\frac{1}{r}\left(1+\frac{r}{R}\right) =\frac{3}{r}-\frac{1}{r}-\frac{1}{R}=\frac{2}{r}-\frac{1}{R}, \end{align}$

implying

(2)

$\displaystyle \sum_{cycl}\frac{R_a}{R_bR_c}=\frac{2R-r}{Rr}.$

Multiplying (1) and (2), we get

$\displaystyle\begin{align} \left(\sum_{cycl}R_a\right)\left(\sum_{cycl}\frac{R_a}{R_bR_c}\right)&\ge (6r)\left(\frac{2R-r}{Rr}\right)\\ &=6\cdot\frac{2R-r}{R}=12-\frac{6r}{R}. \end{align}$

Acknowledgment

This is Dan Sitaru's problem from the Romanian Mathematical Magazine. Solution is by Dan Sitaru.

 

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