Proving $1+\frac{1-\cos x}{x}>0$ for all $x \neq 0$ I proved that
$$1+\frac{1-\cos x}{x}>0$$
for all $x>0$, but I fail to prove the same inequality for $x<0$. I just don't see a way of proving it. I tried proving it at least for $-\pi/2<x<0$: in this case $0<1-\cos x<1$ and then $x+1-\cos x>x$, which is not helpful. Any suggestions?
 A: The case $x>0$ is trivial, since $\frac{1-\cos x}{x} \geq 0$. Let us consider $x <0$. Since $\frac{1-\cos x}{x}$ is an odd function of $x$, your inequality is equivalent to
$$
1-\frac{1-\cos x}{x} >0 \quad\hbox{for all $x>0$}.
$$
By the Mean Value Theorem, given $x>0$ there exists $\xi=\xi(x) \in (0,x)$ such that $1-\cos x = (\sin \xi)x$. As a consequence,
$$
1-\frac{1-\cos x}{x} = 1- \frac{(\sin \xi)x}{x} = 1-\sin \xi.
$$
This immediately gives the weak inequality $\geq 0$. But since you know all the values $\xi$ at which $\sin \xi =1$, you should also be able to prove the strict inequality.
A: As Siminore has stated, $\frac{1-\cos x}{x}$ is an odd function of $x$, thus we only need to prove
$$
1 - \frac{1-\cos x}{x} > 0, \forall x > 0 \tag{1}
$$
I now show another way to prove this. We can simplify (1) to:
$$
x - 1 + \cos x > 0, \forall x > 0
$$
Note that
$$
x - 1 + \cos x = \int_0^x 1 - \sin t \ dt
$$
and
$
f(t) = 1 - \sin t
$ is non-negative when $t > 0$. We conclude that $x - 1 + \cos x > 0$ for $x > 0$.
