Show that $\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$. 
My problem is to show that
  $$\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$$
  for all $x\in\Bbb R$.

I was thinking of first finding the max and then show that its less than $1/\pi$. But it is hard to find it. I get that the series is equal to
$$f(x)=\frac{x-\sin x}{x^2}.$$
Then,
$$f'(x)=\frac{1-\cos x}{x^2}-\frac{2(x-\sin x)}{x^3}=0$$
if and only if
$$2\sin x=x(1+\cos x),$$
which I am unable to solve, appart from the obvious solutions $x=0$ and $x=\pi$. But if $x=\pi$ is the max, then we are done because $f(\pi)=1/\pi$.

Note that $f$ is odd. Thus, it suffices to look at the interval $[0, \infty)$.

Lemma: For $x$ > $\pi$, we have $$\sin(x) > \frac{(\pi^2-x^2)x}{\pi^2+x^2}.$$

Proof: See 
here.
Thus, for $x > \pi$, we have 
$$
\frac{1}x - \frac{\sin(x)}{x^2} < \frac{1}x + \frac{\pi^2-x^2}{x(\pi^2+x^2)} := g(x).$$
For $x > \pi$ (or even $x > 0$), the RHS is monotonically decreasing. Thus, we can conclude that for $x \in (\pi, \infty),$ we have $$f(x) < g(\pi) = \frac{1}{\pi}.$$
Now we will find the maximum of $f$ on $[0, \pi]$. First, we check the endpoints, $$\lim_{x \rightarrow 0} f(x) = 0$$ and $f(\pi) = \frac{1}{\pi}$.
Now we find all the extremas on our desired interval. Taking the derivative and setting it equal to $0$ gives $$2\sin(x)-x \cos(x) - x = -2 \cos(x/2)(x \cos(x/2)-2\sin(x/2)) = 0.$$ The only root of the first factor in our interval is at $x = \pi$. Now we consider $$x \cos(x/2)-2\sin(x/2) = \sqrt{x^2+4}\sin(x/2-\tan^{-1}(2/x)) = 0.$$ The roots of $\sin(\theta)$ are at multiples of $\pi$. Note that $$(x/2-\tan^{-1}(x/2))' = \frac{x^2+8}{2x^2+8}$$ so this function is strictly increasing for $x \ge 0$. Furthermore, $$\lim_{x \rightarrow 0^{+}} \frac{x}2- \tan^{-1}(2/x) = - \frac{\pi}2$$ and $$\frac{\pi}2 - \tan^{-1}(\pi/2) \approx 1$$ so $$\sin(x/2-\tan^{-1}(2/x))$$ has no zeroes in this interval. Thus, the global maximum is at $x = \pi$ which givse $f(\pi) = \frac{1}{\pi}$ as desired. 
