$\cos x -1+\frac{x^2}{2!} \geq 0$ for every $x\in \mathbb{R} $ I am having a problem in showing that $\cos x -1+\dfrac{x^2}{2!} \geq 0$ for every $x\in \mathbb{R}$.
I have tried the following:
I understand that $\dfrac{x^2}{2!}$ is always non negative. But $\cos x-1$ can be negative. So I can't conclude anything about their sum. 
I know that $\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\ldots$ so that $\cos x-1+\dfrac{x^2}{2!}=\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\ldots$ but how do I show that the series in RHS is non-negative? I need some help.
 A: Notice that $f(x)=\cos(x)-1+\frac{x^{2}}{2}$ is even so it suffices to show that it is non-negative on the non-negative reals. Differentiating we get $f'(x)=x-\sin(x)\ge0$ so $f$ is increasing on the non-negative reals. Hence, $f(x)\ge f(0)=0$.
A: By the Maclaurin formula and for $x\in\Bbb R$ there's $\theta\in(0,1)$ such that
$$\cos x=1-\frac{x^2}{2}\cos(\theta x)\ge 1-\frac{x^2}2$$
and the result follows.
A: From $\cos^2x+\sin^2x=1^*$, we have $|\cos x|\le1$, and
$$\color{magenta}{f''(x)}=1-\cos(x)\ge0.$$
The zeroes of $f''$ are isolated, at $x=2k\pi$.
This implies that
$\color{green}{f'(x)}=x-\sin(x)$ is strictly monotone increasing, for all $x$, except at $x=2k\pi$. As $f'(0)=0$, this is the only real root of $f'$.
This implies that $\color{blue}{f(x)}$ is strictly decreasing for $x<0$ and strictly increasing for $x>0$. The global minimum of $f$ is $f(0)=0$, and
$$\cos x-1+\frac{x^2}2\ge0.$$

*$\cos^2x+\sin^2x=(\frac{e^{ix}+e^{-ix}}2)^2+(\frac{e^{ix}-e^{-ix}}{2i})^2=\frac{e^{2ix}+2+e^{-2ix}}4-\frac{e^{2ix}-2+e^{-2ix}}4=1$
A: Perhaps this can be considered a "precalculus" answer.
There are geometric proofs that $|\sin x| \le |x|$.
(for instance, see this: https://math.stackexchange.com/a/75151/1102)
Combine that with $\cos 2x = 1 - 2\sin^2 x$, we get your inequality easily:
$$ \sin^2 (x/2) \le x^2/4 \implies 1 - 2\sin^2 (x/2) \ge 1 - x^2/2 \implies \cos x \ge 1 - x^2/2$$
A: Once you've proven the result for $x \in \mathbb{R}^{+}$ with the above-mentionned geometric evidence, just use the fact that your function is even to get your proof done in $\mathbb{R}^{+} \cup \mathbb{R}^{-}$, which is $\mathbb{R}$.
