Is there an inequality for $\sinh(x)$ which is similar to this inequality $\cosh(x)\leq e^{x^2/2}$

Question

Is there an inequality for $\sinh(x)$ which is similar to this cosh x inequality?

Answer 1

Since: $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) $$ we have: $$ \cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{(2n+1)^2\pi^2}\right)\leq \exp\left(x^2\sum_{n\geq 0}\frac{4}{(2n+1)^2 \pi^2}\right)=e^{x^2/2}. $$ In a similar fashion, from: $$ \sinh(x) = x\prod_{n\geq 1}\left(1+\frac{x^2}{n^2 \pi^2}\right) $$ we get:

$$ \frac{\sinh x}{x}\leq \exp\left(x^2\sum_{n\geq 1}\frac{1}{n^2\pi^2}\right) = e^{x^2/6}.$$

Answer 2

$$ \begin{align} \frac{\sinh(x)}x &=\sum_{k=0}^\infty\frac{x^{2k}}{(2k+1)!}\\ &\le e^{x^2/6}\\ &=\sum_{k=0}^\infty\frac{x^{2k}}{6^k\cdot k!} \end{align} $$ which can be proven by induction and for $k\ge0$, $$ \frac{(2k+3)!}{(2k+1)!}=(4k+6)(k+1)\ge6(k+1)=\frac{6^{k+1}(k+1)!}{6^kk!} $$ Therefore, $$ \frac{\sinh(x)}x\le e^{x^2/6} $$

Answer 3

How about the following: $$\sinh(x) = \frac{e^x-e^{-x}}{2} \leq \frac{e^x}{2}$$

This works since $e^{-x}$ is always positive.

Answer 4

Here is another proof of the inequality $$\frac{\sinh u} u<e^{u^2/6} \tag{$1$}$$ for $u>0$, proved on this page, in different ways, by Jack D'Aurizio and by robjohn:

Let $$h(u):=\frac{\sinh u}u\,e^{-u^2/6} \quad\text{and}\quad h_1(u):=6 u^2 e^{u + u^2/6} \frac{h'(u)}{3 + 3 u + u^2}.$$ Then $$h_1'(u)=-\frac{2u^4 e^{2u}}{(3 + 3 u + u^2)^2}<0$$ for $u>0$. So, $h'<0$ and hence $h$ is decreasing on $(0,\infty)$, down from $h(0+)=1$. So, $h<1$ on $(0,\infty)$, that is, $(1)$ holds for $u>0$.