# 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?

• How on earth is $\cosh$ defined if not by either Taylor series or differential equation? If it is not even defined, then the exercise is meaningless. Commented Feb 8, 2020 at 4:17

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.

• AS mathematics is for 16/17 year olds, so AM-GM inequality is beyond their knowledge. Commented Jan 25, 2020 at 15:13

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

$$(\ln(1+\sqrt2),\cosh(\ln(1+\sqrt2))) = (\ln(1+\sqrt2), \sqrt2)$$

• It's a nice proof only differentiation of hyperbolic functions is not on the syllabus. Commented Jan 25, 2020 at 16:32
• @Paul - As a work around, use $\cosh t = \frac{e^t+e^{-t}}2$. So, its derivative is $(\cosh t)' = \frac{e^t-e^{-t}}2=\sinh t$. Commented Jan 25, 2020 at 16:41

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.

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.