Prove: for $\forall x\ne 0, \cos x < 1 - {x^2\over 2} + {x^4\over 24}$ 
Prove: for $\forall x\ne 0, \cos x < 1 - {x^2\over 2} + {x^4\over 24}$

What I did:
We can prove:
$${\cos x -1 + {x^2\over 2} \over {x^4\over 24}} < 1$$
Lets define:
$f(x) = \cos x -1 + {x^2\over 2}$ and $g(x)= {x^4\over 24}$
By LMVT:
$${{f(x) - f(0)} \over {g(x) - f(0)}} = {f'(y)\over g'(y)} = {{-\sin y + y} \over {4x^3\over 24}} = {{-\sin y + y} \over {x^3\over 6}}\text{ where }y\in(0,x)$$
I tried to show the last expression is smaller than $1$, but without success.
What's the trick? Maybe the $f(x), g(x)$ are wrong?

\begin{align}
f(x) & =1-\dfrac{x^2}{2} + \dfrac{x^4}{24}, & f(0) =1& =\cos0 \\[6pt]
f'(x) & = -x + \frac{x^3}{6}, & f'(0) =0 & =\cos'0 \\[6pt]
f''(x) & = -1 + \frac{x^2}{2}, & f''(0)=-1 & =\cos''0 \\[6pt]
f'''(x) & = x, & f'''(0)  = 0 & = \cos'''0 \\[6pt]
f''''(x) & = 1, & f''''(0)  = 1 & =\cos''''0
\end{align}
We have $f''''(x)\ge\cos''''x$ for all values of $x$, because $f''''$ remains equal to $1$ while $\cos''''=\cos$ oscillates between $1$ and $-1$.
Thus the derivative of $f'''$ is everywhere greater than the derivative of $\cos'''$ and their values at $0$ are equal.  Hence the mean value theorem tells us that $f'''\ge\cos'''$ on $[0,\infty)$ and $f'''\le \cos'''$ on $(-\infty,0]$
Since the derivative of $(f-\cos)''$ is $\ge0$ on $[0,\infty)$ and $\le0$ on $(-\infty,0]$ and $=0$ at $0$, we must have $(f-\cos)''\ge0$ everywhere.  Hence the mean value theorem tells us that the graph of $f-\cos$ is concave upward everywhere.  If it is concave upward everywhere at equal to $0$ at $0$, then it is non-negative everywere.
