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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.

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  • $\begingroup$ Did you use jensen's on log(sin())? $\endgroup$ – Gautam Shenoy Aug 21 '13 at 8:05
  • $\begingroup$ @GautamShenoy you can post this as hint-like answer $\endgroup$ – Norbert Aug 21 '13 at 8:07
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    $\begingroup$ 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. $\endgroup$ – Did Aug 21 '13 at 8:43
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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.

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

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    $\begingroup$ But a<b and c<d can't imply a-b<c-d... $\endgroup$ – A. Chu Aug 21 '13 at 8:28
  • $\begingroup$ But log and log(sin) are both concave hence the convexity/concavity of their difference does not follow. $\endgroup$ – Did Aug 21 '13 at 8:48
  • $\begingroup$ You're right. I flipped the inequality incorrectly. Fortunately the edit I've made suffices. As you suggested. $\endgroup$ – Gautam Shenoy Aug 21 '13 at 8:48
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    $\begingroup$ @GautamShenoy: I should say it does. $\endgroup$ – robjohn Aug 21 '13 at 8:51

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