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

Question

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.

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 08:43

score 3 · Accepted Answer · answered Aug 21 '13 at 08:43

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.

Gautam Shenoy · Answer 2 · 2013-08-21T08:48:00.810

1

Hint: Apply Jensen's on log(sin(x)/x).

edited Aug 21 '13 at 08:48

answered Aug 21 '13 at 08:10

Gautam Shenoy

10,318

1

But a<b and c<d can't imply a-b<c-d... – JSCB Aug 21 '13 at 08:28
But log and log(sin) are both concave hence the convexity/concavity of their difference does not follow. – Did Aug 21 '13 at 08:48
You're right. I flipped the inequality incorrectly. Fortunately the edit I've made suffices. As you suggested. – Gautam Shenoy Aug 21 '13 at 08:48
1

@GautamShenoy: I should say it does. – robjohn Aug 21 '13 at 08:51

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

2 Answers2