# An Inequality with Sines But Not in a Triangle

### Solution

First off, $\displaystyle x\gt 4\Rightarrow x+1\gt x\gt 4\Rightarrow \frac{1}{x+1}\lt \frac{1}{x}\lt \frac{1}{4},\,$ so that

$\displaystyle \frac{2}{x+1}<\frac{2}{x}<\frac{1}{2}.$

And, subsequently,

$\displaystyle 0<\frac{2\pi}{x+1}<\frac{2\pi}{x}<\frac{\pi}{2}.$

Now, applying Jordan's inequality,

$\displaystyle \sin \frac{2\pi}{x}\geq\frac{2}{\pi}\cdot \frac{2\pi}{x}=\frac{4}{x},$

implying

(1)

$\displaystyle a^2\sin \frac{2\pi}{a}\geq 4a$

and also

(2)

$\displaystyle (a+1)^2\sin \frac{2\pi}{a+1}\geq 4(a+1).$

\displaystyle \begin{align} &a^2\sin \frac{2\pi}{a}+(a+1)^2 \sin \frac{2\pi}{a+1}\geq 8a+4\\ &\qquad\qquad \overbrace{\gt}^{AM-GM} 2\sqrt{8a\cdot 4}=8\sqrt{2a}. \end{align}

\displaystyle\begin{align} &\prod \Bigr(a^2\sin \frac{2\pi}{a}+(a+1)^2\sin \frac{2\pi}{a+1}\Bigr)\gt 8^3 \cdot 2\sqrt{abc}\\ &\qquad\qquad =2^9\cdot 2\cdot \sqrt{2\cdot 2^{11}}=2^{10}\cdot 2^6=2^{16}. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later mailed me his solution on a LaTeX file which is greatly appreciated. N. N. Taleb has come independently with the same solution.