Proving $\ln \cosh x\leq \frac{x^2}{2}$ for $x\in\mathbb{R}$

Question

It seems to me that $\ln \cosh x\leq \frac{x^2}{2}$ for $x\in\mathbb{R}$, as suggested by graphing the difference between both functions as well as the fact that the Taylor series expansion of $\ln\cosh x$ at $x=0$ yields $\frac{x^2}{2}-\frac{x^4}{12}+\mathcal{O}(x^6)$. However, how do I prove the bound formally? Using Taylor's Theorem with the remainder seems sort of unwieldy... Any hints?

Answer 1

Since both sides of the inequality are even functions, we need only consider $x\geqslant 0$. Then using

$$\ln \cosh x = \int_0^x \tanh t\,dt$$

one only needs to see that $\tanh t \leqslant t$ for $t \geqslant 0$. Since this inequality is strict for $t > 0$, the original inequality is strict for $x \neq 0$.

Answer 2

One way to do this would be to compare the Maclaurin series expansions of $\cosh x$ and $e^{\frac{x^2}{2}}$:

$\displaystyle\;\;\;\cosh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}\le\sum_{n=0}^{\infty}\frac{x^{2n}}{2^n n!}=e^{\frac{x^2}{2}}$ $\;\;$ since $(2n)!\ge2^n n!$ for all n.