# An Inequality in Triangle, X

### Proof

Since, by the Law of Sines, $\displaystyle\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}= 2R,\;$ where $R\;$ is the circumradius of $\Delta ABC,$

\begin{align} \Delta &= \begin{vmatrix} a& b\cos C & c\cos B\\ b& c\cos A & a\cos C\\ c& a\cos B & b\cos A \end{vmatrix}\\ &=8R^3\begin{vmatrix} \sin A & \sin B\cos C & \sin C\cos B\\ \sin B & \sin C \cos A& \sin A\cos C\\ \sin C & \sin A\cos B & \sin B \cos A \end{vmatrix}\\ &=8R^3\begin{vmatrix} \sin A & \sin (B+C) & \sin C\cos B\\ \sin B & \sin (A+C) & \sin A\cos C\\ \sin C & \sin (A+B) & \sin B\cos A \end{vmatrix}\\ &=8R^3\begin{vmatrix} \sin A & \sin (\pi-A) & \sin C\cos B\\ \sin B & \sin (\pi-B) & \sin A \cos C\\ \sin C & \sin (\pi-C) & \sin B \cos A\\ \end{vmatrix}\\ &=8R^3\begin{vmatrix} \sin A & \sin A & \sin C\cos B\\ \sin B & \sin B & \sin A \cos C\\ \sin C & \sin C & \sin B \cos A\\ \end{vmatrix}\\ &=0. \end{align}

On the other hand,

\displaystyle\begin{align} 0=\Delta &=\begin{vmatrix} a& b\cos C & c\cos B\\ b& c\cos A & a\cos C\\ c& a\cos B & b\cos A \end{vmatrix}\\ &=abc\,\cos^2 A+abc\,\cos^2 C+abc\,\cos^2 A\\ &\;\;\;\;\;- c^3\,\cos A\cos B-a^3\,\cos B\cos C-b^3\,\cos A\cos C\\ &=abc\sum_{cycl} \cos^2 A-\sum_{cycl} a^3 \cos B\cos C\\ &=4RS\sum_{cycl} \cos^2 A-\sum_{cycl} a^3 \cos B\cos C, \end{align}

where $S=[\Delta ABC]\;$ is the area of $\Delta ABC.$ (As is well known, $abc=4RS.)$

Further,

\displaystyle\begin{align} \sum_{cycl} a^3 \cos B\cos C &=4RS\sum_{cycl} \cos^2 A\\ &=4Rrp\sum_{cycl} \cos^2 A\\ &=4Rr\cdot \frac{a+b+c}{2}\sum_{cycl} \cos^2 A \end{align}

so that

\displaystyle\begin{align} \frac{\displaystyle\sum_{cycl} a^3\cos B\cos C}{\displaystyle\sum_{cycl} \cos^2 A}&=2Rr(a+b+c)\\ &=2Rr\cdot 2R\cdot \sum_{cycl} \sin A\\ &=4R^2r\sum_{cycl} \sin A \end{align}

Now using Euler's inequality $R\gt 2r,$

$\displaystyle\frac{\displaystyle\sum_{cycl} a^3\cos B\cos C}{\displaystyle\left(\sum_{cycl} \sin A\right)\left(\sum_{cycl} \cos^2 A\right)}= 4R^2r\ge 16r^3,$

which is the same as the required inequality.

### Acknowledgment

The inequality and the solution have been kindly communicated to me by Dan Sitaru. It was published at the Romanian Mathematical Magazine.

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