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I came across the following assertion and am having trouble justifying it:

If $z$ is a nonzero complex number with $|z| \leq \pi/2$ and $|\sin z| \leq 1/4$, then $ \left| \frac{z}{\sin z} \right| \leq \frac{1/4}{\sin(1/4)} = 1.0104931\ldots $

I would appreciate some help. Thanks.

EDIT: Andrew has pointed out that the above inequality fails if $z = \sin^{-1}(1/4) = 0.25268025\ldots$. After more thought, I figured out that if $z$ is real with $|z| \leq \pi/2$ and $|\sin z| \leq 1/4$, then $|z| \leq \sin^{-1}(1/4)$ and $z / \sin z$ is increasing in $[0,\sin^{-1}(1/4)]$; consequently, $ \left| \frac{z}{\sin z} \right| \leq \frac{\sin^{-1}(1/4)}{1/4} = 1.0107210\ldots $ holds.

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    @HHNT The Maclaurin expansion of $f(z)=z/\!\!\sin z\,$ at the origin has positive coefficients (it follows from the Weierstrass factorization theorem for $\sin z$). Hence the maximum of $f(z)$ in the circle $Q=\{|z|\le z_0=\sin^{-1}(1/4)\}\ $ is attained at $z=z_0$. The function $\sin^{-1}z$ has nonnegative Maclaurin series coefficients at the origin, so the set $\{|\sin z|\le 1/4$, $|z|\le \pi/2\}\ $ is contained in $Q$.2011-07-09

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It is not true. Function $f(x)=\frac x{\sin x}$ is increasing on $(0,\pi/2)$, so function $g(x)=\frac{f(x)}{f(\sin x)} =\frac{x \sin (\sin (x))}{ \sin^2(x)}$ is increasing too since $0<\sin x and $f(0)=1$. So $g(x)\ge1$ on $[0,\pi/2)$. Taking $z=\sin^{-1}(1/4)$ we have $ g(z)=\frac{\sin^{-1}(1/4)}{1/4}\frac{\sin(1/4)}{1/4}>1. $