Prove that $\cosh x > x$

Question

I am attempting to prove that $\cosh x > x$ for all values of $x$. I have proven it by induction but I don't think that would suffice since proof by induction only covers integers.

Also, from a graph you can see clearly that $\cosh x > x$ for all values of $x$ and I am not sure if that would be enough for a proof.

Answer 1

Suppose $x\ge\cosh x$ for some $x\in\Bbb R$. Then all entries in the sequence $u_0:=1,\,u_{n+1}:=\cosh u_n$ are lower bounds on $x$. We need only show this sequence grows without bound. The $u^2$ term in the Taylor series of $\cosh u$ ensures $u_{n+1}\ge\frac12u_n^2$. For $n\ge3$, we can prove by induction that $u_n>5\land u_{n+1}\ge\frac12u_n^2>\frac{5}{2}u_n$.

Answer 2

$cosh(x)=\frac{e^x - e^{-x}}{2}>0$ for all x $cosh(x)>x$ for all $x<0$

$cosh(x)=\frac{e^x + e^{-x}}{2}$, $\frac{dcosh(x)}{dx}=\frac{e^x - e^{-x}}{2}>0$

$cosh(0)=1$, so for $x>0, cosh(x)>1$

So for $0<x<1, cosh(x)>x$,

For $x>1, \frac{dcosh(x)}{dx}=\frac{e^x - e^{-x}}{2}>\frac{e - e^{-1}}{2}>\frac{2.7 - 0.6}{2}=1.05$ This means that$ cosh(x)>x$ for all $x>1$.

$cosh(x)>x$ for all x.