# Proof of $X+\csc{\left(\frac{\pi}{X}\right)}>2\csc{\left(\frac{\pi}{2X}\right)}$

When $X$ is greater than 1, I want to prove that

$X+\csc{\left(\frac{\pi}{X}\right)}>2\csc{\left(\frac{\pi}{2X}\right)}$

where $\csc{(\cdot)}=\frac{1}{\sin{(\cdot)}}$.

Plotting the above expression using computer software, the plot shows the inequality is true.

How to prove it mathematically?

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I suggest that, before doing anything else, you make a change of variables, $Y=1/X$. –  Lubin Sep 27 '13 at 1:59

$\displaystyle{\theta \equiv {\pi \over 2X}\,, \quad 0 < \theta < {\pi \over 2}}$. We have to prove $\displaystyle{{\pi \over 2\theta} + \csc\left(2\theta\right) > \csc\left(\theta\right)}$.

When $\theta \in \left(0, \pi/2\right),\quad$ $\sin\left(\theta\right) > 2\theta/\pi\quad$ such that

$$\color{#ff0000}{\large\quad\csc\left(\theta\right) < \color{#000000}{\pi/\left(2\theta\right)} < \pi/\left(2\theta\right) + \csc\left(2\theta\right)\quad}$$

since $\quad\csc\left(2\theta\right) > 0\quad$ when $\quad\theta \in \left(0, \pi/2\right)$.

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note that if $\displaystyle{\theta \equiv {\pi \over 2X}\,,\quad 0 < \theta < {\pi \over 2}}$. We have to prove $\displaystyle{{\pi \over 2\theta} + \csc\left(2\theta\right) > \color{#ff0000}{2}\csc\left(\theta\right)}$. –  pipi Oct 9 '13 at 2:23

I have a partial result, as follows.

Since $\sin x \le x \, \,\forall \, x \ge 0$, this means that $$-\frac{1}{\sin\frac{\pi}{x}}<-\frac{x}{\pi}$$ .

It was shown in here that $\sin x \ge \frac{x}{x+1}, \space \space\forall x \in \left[0, \frac{\pi}{2}\right]$. This implies that $$\frac{2}{\sin\frac{\pi}{2x}} \le 2+\frac{4x}{\pi}$$.

Hence, $$\frac{2}{\sin\frac{\pi}{2x}}-\frac{1}{\sin\frac{\pi}{x}}\le 2+\frac{4x}{\pi}-\frac{x}{\pi}=2+\frac{3x}{\pi}$$

Now, $2+\frac{3x}{\pi} < x$ if $x>\frac{2\pi}{\pi -3}$, which is about 44.375.

So, I only proved that $X+\csc{\left(\frac{\pi}{X}\right)}>2\csc{\left(\frac{\pi}{2X}\right)}$ if $X>44.4$.

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