# An Inequality with One Tangent and Six Sines

### Solution

Note that, for $\alpha\in \left(0,\displaystyle \frac{pi}{2}\right),\,$ $\tan\alpha\gt\alpha\,$ and $\sin\alpha\lt\alpha.\,$ Thus, suffice it to prove that

$\displaystyle \frac{A}{B+5C}+\frac{B}{C+5A}+\frac{C}{A+5B}\ge\frac{1}{2}.$

As in the proof of Nesbitt's inequality, the above is equivalent to

$\displaystyle \frac{A^2}{A(B+5C)}+\frac{B^2}{B(C+5A)}+\frac{C^2}{C(A+5B)}\ge\frac{1}{2}.$

\displaystyle\begin{align} \sum_{cycl}\frac{A^2}{A(B+5C)}&\ge\frac{\displaystyle \left(\sum_{cycl}A\right)^2}{\displaystyle 6\sum_{cycl}AB}\\ &=\frac{\displaystyle \sum_{cycl}A^2+2\sum_{cycl}AB}{\displaystyle 6\sum_{cycl}AB}\\ &\ge\frac{\displaystyle 3\sum_{cycl}AB}{\displaystyle 6\sum_{cycl}AB}\ge\frac{1}{2}. \end{align}

### Refinement

Given an acute $\Delta ABC.\,$ Prove that

$\displaystyle \frac{\tan A}{\sin B+5\sin C}+\frac{\tan B}{\sin C+5\sin A}+\frac{\tan C}{\sin A+5\sin B}\ge 1.$

Indeed, using the AM-GM inequality, we obtain

\displaystyle\begin{align} \sum_{cycl}\frac{\tan A}{\sin B+5\sin C} &\ge 3\sqrt[3]{\prod_{cycl}\frac{\tan A}{\sin B+5\sin C}}\\ &=3\frac{\sqrt[3]{\tan A\tan B\tan C}}{\displaystyle\sqrt[3]{\prod_{cycl}(\sin B+5\sin C)}}. \end{align}

However, $\displaystyle\prod_{cycl}\tan A=\sum_{cycl}\tan A\ge 3\sqrt[3]{\prod_{cycl}\tan A},\,$ implying $\displaystyle \prod_{cycl}\tan A\ge 3\sqrt{3}.\,$ To continue,

\displaystyle \begin{align} \sqrt[3]{\prod_{cycl}(\sin B+5\sin C)}&\le\frac{\displaystyle \sum_{cycl}(\sin B+5\sin C)}{3}\\ &=2\sum_{cycl}\sin A\le 3\sqrt{3}. \end{align}

Combining the latest results,

$\displaystyle \sum_{cycl}\frac{\tan A}{\sin B+5\sin C}\ge 3\frac{\sqrt{3}}{3\sqrt{3}}=1.$

Equality is attained for $\displaystyle A=B=C=\frac{\pi}{3}.$

### Acknowledgment

The problem (from the Romanian Mathematical Magazine) has been posted by Dan Sitaru at the CutTheKnotMath facebook page, Dan later communicated a solution by Soumava Chakraborty. Marian Dinca came up with the refinement of the original inequality.