# An Inequality with Inradius and Excenters

### Solution 1

We know that (withe $R\,$ the circumradius)

$\displaystyle \begin{array}{ccc} II_a=4R\sin\frac{A}{2} & II_b=4R\sin\frac{B}{2} & II_c=4R\sin\frac{C}{2}\\ I_aI_b=4R\cos\frac{C}{2} & I_bI_c=4R\cos\frac{A}{2} & I_cI_a=4R\cos\frac{B}{2} \end{array}$

and also $\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le\frac{1}{r^2}.\,$ The required inequality is equivalent to

\displaystyle \begin{align} &\frac{1}{16R^2}\sum_{cycl}\csc^2\frac{A}{2}+\frac{1}{16R^2}\sum_{cycl}\sec^2\frac{A}{2}\le\frac{1}{4r^2}\;&\Leftrightarrow\\ &\frac{1}{16R^2}\left(\sum_{cycl}\left(\csc^2\frac{A}{2}+\sec^2\frac{A}{2}\right)\right)\le\frac{1}{4r^2}\;&\Leftrightarrow\\ &\frac{1}{16R^2}\left(\sum_{cycl}\left(\tan\frac{A}{2}+\cot\frac{A}{2}\right)^2\right)\le\frac{1}{4r^2}\;&\Leftrightarrow\\ &\frac{1}{16R^2}\left(4\csc^2A+4\csc^2B+4\csc^2C\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le\frac{1}{4r^2}. \end{align}

### Solution 2

We know that (withe $R\,$ the circumradius)

\displaystyle\begin{align} II_a&=a\sec\frac{A}{2}=2R\sin A\sec\frac{A}{2}=4R\sin\frac{A}{2}\cos\frac{A}{2}\sec\frac{A}{2}\\ &=4R\sin\frac{A}{2},\\ I_aI_b&=\csc\frac{A}{2}=2R\sin A\csc\frac{A}{2}=4R\sin\frac{A}{2}\cos\frac{A}{2}\csc\frac{A}{2}\\ &=4R\cos\frac{A}{2}. \end{align}

It follows that

\displaystyle \begin{align} \sum_{cycl}\frac{1}{II_a^2}+\sum_{cycl}\frac{1}{I_aI_b^2} &=\frac{1}{16R^2}\sum_{cycl}\left(\frac{1}{\sin^2\frac{A}{2}}+\frac{1}{\cos^2\frac{A}{2}}\right)\\ &=\frac{1}{16R^2}\sum_{cycl}\frac{1}{\sin^2\frac{A}{2}\cos^2\frac{A}{2}}\\ &=\sum_{cycl}\frac{1}{\displaystyle \left(4R\sin\frac{A}{2}\cos\frac{A}{2}\right)^2}\\ &=\sum_{cycl}\frac{1}{\displaystyle \left(2R\sin A\right)^2}=\sum_{cycl}\frac{1}{a^2}=\frac{\displaystyle \sum_{cycl}a^2b^2}{a^2b^2c^2}\\ &\le\frac{4R^2s^2}{a^2b^2c^2}=\frac{4R^2s^2}{16R^2r^2s^2}=\frac{1}{4r^2}, \end{align}

where at the penultimate step we used Goldstone's inequality: $\displaystyle \sum_{cycl}a^2b^2\le 4R^2s^2,\,$ with $s\,$ the semiperimeter of $\Delta ABC.$

### Acknowledgment

I am grateful to Dan Sitaru for kindly posting a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Soumava Chakraborty.