Prove that: $\cos(x) -1 < -\frac{x^2}{2} + \frac{x^4}{24}$ 
Prove that:
$$\cos(x) -1 < -\frac{x^2}{2} + \frac{x^4}{24}$$ for $x \ne 0$

I need to prove this using Cauchy's mean value theorem.
What I did:
$f(x) = \cos(x) -1$
$$g(x) = -\frac{x^2}{2} + \frac{x^4}{24}$$
If I can show that:
$\frac{f(x)}{g(x)}\lt 1$ that would prove the original inequality
so,
$$\frac{f(x)}{g(x)} = \frac{f(x) - f(0)}{g(x) - g(0)} = \frac{f'(c)}{g'(c)}$$ for $0\lt c \lt x$
$$\dfrac{f'(c)}{g'(c)} = \dfrac{-\sin(c)}{-c + \frac{c^3}{6}}$$
My problem is the following expression is not always less then 1! (for example $c=2.4$)
What am I doing wrong?

I think it will be easier to show that: $$ \cos x - 1 + \frac{x^2}2 < \frac{x^4}{24} $$
Define $f(x) = \cos x - 1 + \frac{x^2}2$, $g(x) = \frac{x^4}{24}$.
Now using the Cauchy MVT: $$ \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{6(c-\sin c)}{c^3} $$
Now just show that $ h(x) = \frac{6(x-\sin x)}{x^3} $ is bounded above globably by $1$, and you'll get what you need.
