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 $f(\pi)=1/\pi$.
Using $f(x)=\frac{x-\sin(x)}{x^2}$, the inequality is equivalent to:
$$
\frac{x-\sin(x)}{x^2}≤\frac{1}{\pi}\iff x\left(1-\frac{x}{\pi}\right)≤\sin(x)
$$
Thus, define $g(x):=x\left(1-\frac{x}{\pi}\right)$. 

Note that $g(x)=g(\pi-x)$ and $\sin(x)=\sin(\pi-x)$. Therefore it suffices (by symmetry) to prove the inequality for $x\in\left(-\infty;\frac{\pi}{2}\right]$.

Firstly, we prove it for $x\in\left[0;\frac{\pi}{2}\right]$: 

We have $g(0)=\sin(0)=0$, thus it is sufficient to prove, that $g'(x)≤\sin'(x)$ for $x\in\left[0;\frac{\pi}{2}\right]$. Now: $g'(x)=1-\frac{2x}{\pi}$ and $\sin'(x)=\cos(x)$. So it remains to prove: 
$$
1-\frac{2x}{\pi}≤\cos(x)
$$
Since $\cos''(x)=-\cos(x)≤0\space\forall x\in\left[0;\frac{\pi}{2}\right]$, $\cos(x)$ is concave on the interval. As $g'(0)=\cos(0)=1$ and $g'\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}\right)=0$, $g'(x)$ connects two points on the graph of $\cos(x)$, and together with the concavity, this implies the above inequality.

Secondly, we prove it for $x\in\left(-\infty;0\right]$:

By the same argument as above, we just have to show, that on this interval $g'(x)≥\sin'(x)\iff 1-\frac{2x}{\pi}≥\cos(x)$ holds. This is true because $1-\frac{2x}{\pi}≥1≥\cos(x)$ for all $x\in\left(-\infty;0\right]$

So the inequality is indeed true.
That is o.k. for obtaining which solution get max and which min use second derivative test: if $f^"(a)&lt;0$ then $a$ get max and if $f^"(a)&gt;0$ then $a$ get min.