How to show that $x \sin \frac{\pi}{x} > \pi \cos \frac{\pi}{x}$ for $x \in (1, \infty)$? How could I prove that $x \sin \frac{\pi}{x} > \pi \cos \frac{\pi}{x}$ for $x \in (1, \infty)$? 
Dividing both sides through by $x \sin \frac{\pi}{x}$ and letting $y = \frac{\pi}{x}$ gives the inequality $1> y \cot y$, if $y \in (0,\pi)$, but then I don't know how to go ahead and actually prove that?
 A: As you noticed, the inequality is equivalent to:
$$ \forall z\in(0,1),\qquad \frac{\sin(\pi z)}{\pi z} > \cos(\pi z) \tag{1}$$
or to:
$$\forall z\in(0,1),\qquad \prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) > \prod_{n\geq 1}\left(1-\frac{4z^2}{(2n-1)^2}\right)\tag{2} $$
that is trivial.
A: Your substitution allows to prove
$$
\frac{\pi}{y}\sin y>\pi\cos y
$$
for $0<y<\pi$, that is,
$$
\sin y> y\cos y
$$
If $\pi/2\le y<\pi$, the statement is obvious. For $0<y<\pi/2$ it becomes
$$
y<\tan y
$$
which is well known. If you don't know it, consider
$$
f(y)=y-\tan y
$$
so $f(0)=0$ and
$$
f'(y)=1-1-\tan^2y<0
$$
Thus $f$ is decreasing on $(0,\pi/2)$.
A: If $x\le2$, it is true. If $x>2$, the inequality is equivalent to
$$ \tan y>y, \forall y\in(0,\pi/2). $$
Let $f(y)=\tan y-y$. Then $f'(y)=\sec^2y-1>0$ and hence $f(y)$ is increasing in $(0,\pi/2)$. So $f(y)>f(0)=0$ or
$$ \tan y>y, \forall y\in(0,\pi/2). $$
Done.
A: Using the substitution $y=\pi/x$ works well.
Then, for $y\in (0,\pi)$, we have 
$$y\sin(y)>0 \tag 1$$  
Integrating $(1)$, we obtain
$$\begin{align}
0&<\int_0^y t\sin(t)\,dt\\\\
&=\sin(y)-y\cos(y)
\end{align}$$
whereupon we find $\sin(y)>y\cos(y)$ for $0<y<\pi$.  Substituting back yields
$$x\sin(\pi/x)>\pi\cos(\pi/x)$$
for $x\in (1,\infty)$.  And we are done!
