Showing that $1 - \frac{x^2}2\leq\cos x$, $\forall x \in \mathbb{R}$ 
Show that
  $$\displaystyle1 - \frac{x^2}2\leq\cos x\quad\forall x \in \mathbb{R}$$

Let $f(x) = \cos x - 1 + \frac{x^2}2$; then we need to show that $f(x) \geq 0\quad\forall x \in \mathbb{R}$.
Since
$$\lim_{x \to \pm\infty} f(x) = +\infty$$
and
$$\operatorname{im}f'' = [0, 2]$$
(as $f''(x) = 1 -\cos x$), it remains to show that $x = 0$ is a minimum for $f(x)$, but I cannot do that because I would need to solve $f'(x) = x -\sin x = 0$, which I cannot.
I think it would suffice to show that plugging in $x = 0$ we have that $f'(x) = 0$, and combining it with the fact that the function $f$ is convex, we would be done.  
Is it enough? What's a better proof?
EDIT: It just occurred to me that there is a way to solve $f'(x) = 0$. Just observe that $f'(x)$ is an odd function and that $f''(x) \geq 0$ implies that $f'(x)$ is monotonically increasing. Since an odd function, if defined at $x = 0$, must be $0$ there, $x = 0$ is a solution to the equation and it's the only one.
 A: Since $\cos x$ and $1 - \frac{x^2}{2}$ are even functions it suffices to prove $1 - \frac{x^2}{2} \le \cos x$ for all $x \ge 0$. For all $x \ge 0$,
$$\sin x = \int_0^x \cos t\, \mathrm{d}t \le \int_0^x 1\, \mathrm{d}t = x.$$
Hence, for all $x \ge 0$,
$$\cos x = 1 + \int_0^x -\sin t\, \mathrm{d}t \ge 1 + \int_0^x -t\, \mathrm{d}t = 1 - \frac{x^2}{2}.$$
A: let $$f(x)=\cos(x)-1+\frac{x^2}{2}$$ then we get $$f'(x)=-\sin(x)+x$$ this equation has only the solution $$x=0$$ thus by inserting in $$f^{(iv)}(x)=\cos(x)$$ we get $$1>0$$ and thus we have a local minimum with the limites derived above we obtain that is a global one.
A: Your idea may very well be more elegant, but for $0<x<\pi,$
we have
$$\frac{d}{dx}(x-\sin x) = 1 - \cos x > 0,$$
hence $f(x) = x-\sin x > 0$ since $f(0)= 0.$
For $x \geq \pi,$ we have
$$x - \sin x \geq \pi - 1 > 0.$$
So $x-\sin x > 0$ for all $x > 0.$ 
And by symmetry, $x-\sin x < 0$ for all $x < 0.$
A: Notice that 
$$f'(x) =  x-\sin(x).$$
If $x \geq 0$, then $x\geq \sin(x)$ so that $f'(x) \geq 0$, if $x\leq 0$ then $f'(x)\leq 0$ so we have a global minimum at $x=0$ where $f(x)=0$. Thus $f(x) \geq 0$ for al $x\in \Bbb R$.
