Elementary proof of a cotangent inequality 
Let $0<x<\pi/2$. Then
  $$ \cot{x} > \frac{1}{x}+\frac{1}{x-\pi}. $$

(This is still true for $-\pi<x<0$, but the given range is the one I'm concerned about.)
Is there an elementary proof of this? The local inequality $\cot{x}<1/x$ is easy to prove since it is equivalent to $x<\tan{x}$, which even has a simple geometric proof.

One can massage the Mittag-Leffler formula
$$ \cot{x} = \frac{1}{x} + \sum_{n \neq 0} \frac{1}{x-n\pi} + \frac{1}{n\pi}  $$
into
$$ \cot{x} = \frac{1}{x} + \frac{1}{x-\pi} + \sum_{n=1}^{\infty} \frac{1}{z- (n+1)\pi}+\frac{1}{z+\pi n}, $$
and the terms in the second sum are all positive in the range considered since
$$ \frac{1}{z- (n+1)\pi}+\frac{1}{z+\pi n} = \frac{\pi-2z}{n(n+1)\pi^2+z(\pi-z)}>0, $$
but this is rather heavyweight for such a simple-looking inequality.

An equivalent formulation is
$$ \tan{y} > \frac{1}{\pi/2-y}-\frac{1}{\pi/2+y} = \frac{8y}{\pi^2-4y^2} $$
for $0<y<\pi/2$, if one desires more symmetry.
 A: Note that on the interval $[-\pi,\pi]$ $$0 \leq \operatorname{sinc}(x)=\frac{\sin(x)}{x}=\frac{\sin(x/2)\cos(x/2)}{x/2} \leq \cos(x/2)$$ with equality only in $\{-\pi,0,\pi\}$. Then on $(0, \pi)$ 
$$\begin{eqnarray}
\operatorname{sinc}^2(x) + \operatorname{sinc}^2(x-\pi) &< &\cos^2(x/2) + \cos^2((x-\pi)/2)\\& =& \cos^2(x/2) + \sin^2(x/2)\\& =& 1 \end{eqnarray}.$$ Dividing both sides by $\sin^2(x)$ on the same interval and noting that $\sin^2(x)=\sin^2(x-\pi)$ $$\frac1{x^2}+\frac1{(x-\pi)^2}<\frac1{\sin^2(x)}.$$ Then for $x\in(0,\pi/2)$ $$\frac1x + \frac1{x-\pi} = \int_x^{\pi/2}\left(\frac1{t^2}+\frac1{(t-\pi)^2}\right)\mathrm{d}t < \int_x^{\pi/2}\frac{\mathrm{d}t}{\sin^2(t)} = \cot(x).$$
A: We need to prove that
$$\frac{1}{\tan{x}}>\frac{\pi-2x}{x(\pi-x)}$$ or
$$\pi x-x^2>(\pi-2x)\tan{x}$$ or 
$$x^2-(2\tan{x}+\pi)x+\pi\tan{x}<0$$ or
$$\tan{x}+\frac{\pi}{2}-\sqrt{\tan^2x+\frac{\pi^2}{4}}<x<\tan{x}+\frac{\pi}{2}+\sqrt{\tan^2x+\frac{\pi^2}{4}},$$
for which it's enough to prove that $f(x)>0$, where
$$f(x)=x-\tan{x}-\frac{\pi}{2}+\sqrt{\tan^2x+\frac{\pi^2}{4}},$$
which is easy.
Indeed,
$$f'(x)=1-\frac{1}{\cos^2x}+\frac{\tan{x}}{\cos^2x\sqrt{\tan^2x+\frac{\pi^2}{4}}}=$$
$$=\frac{\sin{x}\left(1-\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)}{\cos^3x\sqrt{\tan^2x+\frac{\pi^2}{4}}}=$$
$$=\frac{\sin{x}\left(1-\sin^4x-\frac{\pi^2}{4}\sin^2x\cos^2x\right)}{\cos^3x\sqrt{\tan^2x+\frac{\pi^2}{4}}\left(1+\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)}=$$
$$=\frac{\sin{x}\cos^2x\left(1+\sin^2x-\frac{\pi^2}{4}\sin^2x\right)}{\cos^3x\sqrt{\tan^2x+\frac{\pi^2}{4}}\left(1+\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)}=$$
$$=\frac{\left(\frac{\pi^2}{4}-1\right)\sin{x}\left(\frac{2}{\sqrt{\pi^2-4}}-\sin{x}\right)\left(\frac{2}{\sqrt{\pi^2-4}}+\sin{x}\right)}{\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\left(1+\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)},$$
which gives $x_{max}=\arcsin\frac{2}{\sqrt{\pi^2-4}}$ and since $$\lim_{x\rightarrow0^+}f(x)=\lim_{x\rightarrow\frac{\pi}{2}^-}f(x)=0,$$
we are done!
