An Inequality in Triangle, X

Problem

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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