<s>[INST] 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?
 [/INST] This solution is way too long but can be tightened up and only uses Calc I methods.
First, the only solution of $\sin x = x$ is $x = 0$. 
If there existed a second point $x$ with $\sin x = x$ the mean value theorem would provide  a point $c$ in between $0$ and $x$ with $\cos c = 1$. This means $c$ is a multiple of $2\pi$, so that $|x| \ge 2\pi$. Since $|\sin x| = 1$ this is impossible.
Second, the only solution of $\cos x = 1 - \dfrac{x^2}{2}$ is $x = 0$.
Define $f(x) = 1 - \dfrac{x^2}{2} - \cos x$. If there existed a second point with $f(x) = 0$, the mean value theorem would provide a point $c$ in between $0$ and $c$ with $f'(c) = 0$. This point satisfies $\sin c = c$, which is impossible since $c \not = 0$. This means that $f$ has no zeroes other than $x = 0$. 
Third, $1 - \dfrac{x^2}{2} < \cos x$ for all $x \not= 0$.
Define $f$ as above. Since the only zero of $f$ is at $x=0$, it suffices to show that $f(x)$ is negative in a neighborhood of $0$. According to the intermediate value theorem, $f$ can then never be positive.  You can use the first derivative test to show that $f$ has a local maximum at $x = 0$ and is thus negative in a neighborhood of $0$.
Fourth, the only solution of $\sin x = x - \dfrac{x^3}{3}$ is $x=0$.
If there existed a point $x \not= 0$ satisfying this, the mean value theorem would provide a point $c$ in between $0$ and $x$ with $\cos c = 1 - \dfrac{c^2}{2}$, contrary to the above result.
Finally, $\cos x < 1 - \dfrac{x^2}{2} - \dfrac{x^4}{24}$ for all $x \not= 0$.
Define $g(x) = \cos x - 1 + \dfrac{x^2}{2} - \dfrac{x^4}{24}$. Argue as above. Show that $g$ has only one zero, and that $g$ must be negative in a neighborhood of  $x=0$. Then you're done!
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