# Show that $(\sin x/x)^n\geq(\sin x_1\sin x_2…\sin x_n)/(x_1 x_2 …x_n)$.

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

$L.H.S\geq(\sin x_1+\sin x_2+...+\sin x_n)/(x_1+x_2+ ...+x_n)$.

• Did you use jensen's on log(sin())? – Gautam Shenoy Aug 21 '13 at 8:05
• @GautamShenoy you can post this as hint-like answer – Norbert Aug 21 '13 at 8:07
• One could rather use the fact that $u:x\mapsto\log(\sin(x)/x)$ is concave by checking that $u''(x)$ has the sign of $\sin^2(x)-x^2\leqslant0$, and apply Jensen to that. – Did Aug 21 '13 at 8:43

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.