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?

We all know that
$$\cos x=1-{x^2\over2}+{x^4\over24}-{x^6\over 720}+\ldots\ .$$
When $x^2<30$ then the terms of this  alternating series decrease in absolute value after ${x^4\over24}$, and this implies
$$\cos x<1-{x^2\over2}+{x^4\over 24}\qquad(x^2<30)\ .$$
When $x^2\geq30$ one has
$$1-{x^2\over2}+{x^4\over24}=1+{x^2\over24}(x^2-12)>1\geq\cos x$$
as well, and this completes the proof.
