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

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

A: $$
\begin{align}
\frac{\sinh(x)}x
&=\sum_{k=0}^\infty\frac{x^{2k}}{(2k+1)!}\\
&\le e^{x^2/6}\\
&=\sum_{k=0}^\infty\frac{x^{2k}}{6^k\cdot k!}
\end{align}
$$
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,
$$
\frac{\sinh(x)}x\le e^{x^2/6}
$$
A: 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. 
A: 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 $u>0$.
