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?
 A: 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!
A: let $f(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\cos x$. Clearly $f$ is even, so we only have to show that $f(x)>0$ for $x>0$.
Now
$$\eqalign{f'(x)&=-x+\frac{x^3}{6}+\sin  x\cr
f''(x)&=-1+\frac{x^2}{2}+\cos  x\cr
f^{(3)}(x)&=x-\sin  x\cr
f^{(4)}(x)&=1-\cos x\cr
}$$
$f^{(4)}$ is non-negative on $[0,+\infty)$ and does not vanish on any sub-interval, so it is increasing, and with $f^{(3)}(0)$ we conclude that $f^{(3)}>0$ on $(0,+\infty)$. So, $f''$ is also increasing, and from $f''(0)$ we conclude that $f''>0$ on $(0,+\infty)$. So, $f'$ is also increasing, but $f'(0)=0$ so, $f$ itself is increasing, and the fact that $f(0)=0$ implies that $f(x)>0$ for $x>0$, and the desired conclusion follows.
A: 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.
A: \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.
