Prove $~\cosh x > x~,~~~ \forall ~~x~$ Exercise $5.1$, question 13 of the AQA Further Mathematics for core year 1 and AS has the following question:
$a) ~~$Sketch the curve $~y = \cosh x~$ and the line $~y = x~$ on the same axes.  Prove that $~\cosh x > x~,~~ \forall ~x~$.
This exercise is before hyperbolic differentiation, Maclaurin series or Taylor series.  I can't see how this proof can be shown algebraically using AS level mathematics. The sketch is given as the answer, would I be correct in assuming that there isn't a simple algebraic proof and the authors have decided the sketch is a proof.
I'm also working on part $b)$ Prove that the point on the curve $~y = \cosh x~$ which is closest to the line $~y = x~$ has coordinates $~\left(\ln\left(1 + \sqrt{2}\right), ~\sqrt{2}\right)~$.  
Suggestions would be welcome.
An internet search hasn't helped.  Any ideas?  
 A: Hint: if $x < 1 \implies \cosh x \ge 1 > x $ by AM-GM inequality. If $x \ge 1 \implies e^x -2x > 0$. Few details have been left out for you to fill out.
A: Let $f(x)=\cosh x -x$ and evaluate
$$f'(x)=\sinh x -1 \>\>\bigg\{ \begin{array}{c} \ge 0, \>\>\>x\ge\ln(1+\sqrt2), \\ <0, \>\>\> x<\ln(1+\sqrt2) \ \end{array}$$
where $\sinh^{-1}t=\ln(t+\sqrt{1+t^2})$ is used. So, $f(x)$ strictly decreases for $x<\ln(1+\sqrt2) $ and strictly increases for $x>\ln(1+\sqrt2)$. Then, 
$$f(x) > f(\ln(1+\sqrt2))
=\cosh(\ln(1+\sqrt2))-\ln(1+\sqrt2)=\sqrt2-\ln(1+\sqrt2)>0$$
Thus, 
$$\cosh x -x>0$$
Note that the closest point to the line is 
$$(\ln(1+\sqrt2),\cosh(\ln(1+\sqrt2))) = (\ln(1+\sqrt2), \sqrt2)$$
A: Consider that
$$\cosh(x) \equiv \frac{e^{x} + e^{-x}}{2}$$
Now just remember that the exponential $e^{x}$ (as well as $e^{-x}$) is always positive. Also you have a SUM between the two.
Argue a bit on it and I believe you are easily done.
A: The easiest way to define $\cosh$ is to define it in terms of $\exp$, but that still requires a definition of $\exp$. The problem is, there is no simple way to define $\exp$. Every definition of $\exp$ must pass through a significant amount of real analysis. However, there is a way if the syllabus in question cheats and assumes without justification the following non-trivial facts:

$\exp(x+1) > 2·\exp(x) > 0$ for every real $x$.
$\exp(x) ≥ 1+x$ for every real $x$.
$\exp(-x) = 1 / \exp(x) > 0$ for every real $x$.

$\cosh(x) = (\exp(x)+\exp(-x))/2$ $> (2·\exp(x-1)+0)/2$ $= \exp(x-1) ≥ x$ for every real $x$.
In case it is not clear, there are many other possible ways to derive the same result, but every single way will definitely rely on some non-trivial fact about $\exp$ or a related function.
