TITLE: Show that $(\sin x/x)^n\geq(\sin x_1\sin x_2...\sin x_n)/(x_1 x_2 ...x_n)$.
QUESTION [5 upvotes]: Suppose $0<x_i<\pi$ for $i=1,2,...n$ and $x=(x_1+x_2+...+x_n)/n.$
Show that $(\sin x/x)^n\geq(\sin x_1\sin x_2...\sin x_n)/(x_1 x_2 ...x_n)$.

By Jensen inequality, I showed that
$L.H.S\geq(\sin x_1+\sin x_2+...+\sin x_n)/(x_1+x_2+ ...+x_n)$.
But I don't know what to do then. Please help, Thanks.

REPLY [3 votes]: First note that since $x\ge\sin(x)\ge0$ on $[0,\pi]$,
$$
\begin{align}
\frac{\mathrm{d}^2}{\mathrm{d}x^2}\Big(\log(\sin(x))-\log(x)\Big)
&=\frac1{x^2}-\csc^2(x)\\[6pt]
&\le0
\end{align}
$$
Therefore,
$$
f(x)=\log\left(\frac{\sin(x)}{x}\right)
$$
is concave. Jensen's inequality says that
$$
f\left(\overline{x_i}\right)\ge\overline{f(x_i)}
$$
which is $\frac1n$ times the log of the given inequality.