Proof of $\frac{1}{\sin{(\frac{\pi}{2x})}}<\frac{2x}{\pi}+1$ I want to show that 
\begin{equation}
\frac{1}{\sin{(\frac{\pi}{2x})}}<\frac{2x}{\pi}+1
\end{equation}
for any positive integer $x$.
Seems that it is related to the well-known inequality
\begin{equation}
\frac{1}{\sin{(\frac{\pi}{2x})}}>\frac{2x}{\pi}.
\end{equation}

$$\sin(x) = \sum_{n=0}^{\infty}{\dfrac{x^{2n+1}(-1)^{n}}{(2n+1)!}}$$ $$\sin(\pi/2x) = \sum_{n=0}^{\infty}{\frac{\pi^{2n+1}(-1)^{n}}{(2x)^{2n+1}(2n+1)!}}$$
The inequality obviously holds for $x=1$ and $x = 2$.  For all $x \ge 3$, the above sum is a diminishing alternating series that starts with $\dfrac\pi{2x} - \dfrac{\pi^3}{8x^3}$, and the remainder of the series sums to a positive number.  So $\sin{\dfrac{\pi}{2x}} \gt \dfrac\pi{2x} - \dfrac{\pi^3}{8x^3}$
I want to swap a different term for $\frac{\pi^3}{8x^3}$ in my inequality, so I need to prove it is larger than $\frac{\pi^3}{8x^3}$
$\dfrac{\pi^2}{4x^2+2x\pi} -\dfrac{\pi^3}{8x^3}  = \dfrac{\pi^2 8x^3 - \pi^3(4x^2+2\pi)}{8x^3(4x^2+2x\pi)} = \pi^2\dfrac{4x^3 - \pi(2x^2+\pi)}{4x^3(4x^2+2x\pi)}$
Since the $\pi^2$ and the denominator are both positive, to determine the sign of the difference of the terms I just need to look at the sign of $4x^3 - \pi(2x^2 + \pi)$ $\gt 4x^3 - 4(2x^2 + 4)$, which is positive for $x \ge 3$.
Now I can say (for $x >= 3$) $\dfrac{\pi^2}{4x^2+2x\pi} \gt \dfrac{\pi^3}{8x^3}$, and $0 > \dfrac{\pi^3}{8x^3} - \dfrac{\pi^2}{4x^2+2x\pi}$, which when added to the inequality above gives $\sin{\dfrac{\pi}{2x}} \gt \dfrac\pi{2x} - \dfrac{\pi^2}{4x^2+2x\pi}$ $= \dfrac{\pi(2x+\pi) - \pi^2}{2x(2x + \pi)} = \dfrac{\pi}{2x+\pi}$
$\sin{\dfrac{\pi}{2x}} \gt \dfrac\pi{2x+\pi}$, and both sides are positive, so $\dfrac1{\sin{\dfrac{\pi}{2x}}} < \dfrac{2x}\pi + 1$
