Showing inequality: $pe^{s(1-p)} + (1-p)e^{-sp} \leq e^{s^2 (3/2) p }$ I've been struggling with this inequality for a while. I tried expanding the exponentials into their taylor series, and I tried using convexity but I haven't been able to find a way to show this. 
How can I show that for $0 \leq p \leq .5$, $s \in [0,p]$ we have that: 
$$ pe^{s(1-p)} + (1-p)e^{-sp} \leq e^{s^2 (3/2)  p } $$
Thank you!
 A: We show that, if $0\leq p\leq 1$ and $0\leq s\leq p$, then
$$p\mathrm{e}^{(1-p)s}+(1-p)\mathrm{e}^{-s}\leq \mathrm{e}^{\tfrac{1}{2}ps^2}\text{.}\tag{1}$$
Let $q=1-p$. From Lagrange's form of the remainder for Taylor's theorem, there exists a $\sigma\in[0,p]$ such that
$$\ln(q+p\mathrm{e}^{s})=ps +\tfrac{ps^2}{2} \tfrac{q\mathrm{e}^{\sigma}}{(q+p\mathrm{e}^{\sigma})^2}\text{.}$$
Note that the function
$$x\mapsto \frac{qx}{(q+px)^2}$$
is monotone on $$0\leq x \leq \tfrac{q}{p}\text{.}$$ We consider two cases 1. $p\leq \tfrac{1}{3}$ and 2. $p\geq \tfrac{1}{4}$. 


*

*If $p\leq \tfrac{1}{3}$, then from the inequality $$\mathrm{e}^u\leq 1+\frac{u}{1-\tfrac{1}{2}u}$$ we find $$\mathrm{e}^{\sigma}\leq\mathrm{e}^{1/3}\leq 1+\tfrac{3}{5}\leq 2$$ and $$\frac{q}{p}\geq 2\text{.}$$ Consequently, $\frac{q\mathrm{e}^{\sigma}}{(q+p\mathrm{e}^{\sigma})^2}$ is monotone in $\sigma$, giving $$\begin{split}\frac{q\mathrm{e}^{\sigma}}{(q+p\mathrm{e}^{\sigma})^2}&\leq \frac{q\mathrm{e}^p}{(q+p\mathrm{e}^p)^2}\\ &\leq \frac{q\mathrm{e}^p}{(q+p)^2}\\ &= (1-p)\mathrm{e}^p\\
&\leq (1-p)(1+\tfrac{p}{1-\tfrac{1}{2}p})\\
&=1- \frac{\tfrac{1}{2}p^2}{1-\tfrac{1}{2}p}\\
&\leq 1\text{.}\end{split}$$

*If $p\geq \tfrac{1}{4}$, then $\frac{q\mathrm{e}^{\sigma}}{(q+p\mathrm{e}^{\sigma})^2}$ is bounded above by its value at $\mathrm{e}^{\sigma}=\tfrac{q}{p}$: $$\begin{split}\frac{q\mathrm{e}^{\sigma}}{(q+p\mathrm{e}^{\sigma})^2}&\leq \frac{1}{4p}\\ &\leq 1\text{.}\end{split}$$


In either case we have $$\frac{q\mathrm{e}^{\sigma}}{(q+p\mathrm{e}^{\sigma})^2}\leq 1\text{.}$$ Therefore
$$\ln(q+p\mathrm{e}^{s})\leq ps +\tfrac{ps^2}{2}$$
from which (1) is immediate.
Let $s=pt$. Then
$$\lim_{p\to 0}\frac{\ln(q+p\mathrm{e}^{pt})- p^2t}{\tfrac{1}{2}p^3t^2}=1\text{,}$$
so this inequality can't be sharpened.
An older version of this answer showed that if $0\leq p\leq 1$ and $0\leq s\leq 1$, then
$$p\mathrm{e}^{(1-p)s}+(1-p)\mathrm{e}^{-s}\leq \mathrm{e}^{p(1-p)s^2}\text{.}$$
Let $q=1-p$. From the integral form of the remainder for Taylor's theorem, we have
$$\ln(q+p\mathrm{e}^{s})=ps +pqs^2 \int_0^1 \frac{\mathrm{e}^{st}}{(q+p\mathrm{e}^{st})^2} (1-t)\mathrm{d}t\text{.}$$
From the elementary inequality 
$$1-t\leq \mathrm{e}^{-t}$$
and since $0\leq s \leq 1$ and $t\geq 0$ imply
$$\mathrm{e}^{-t}\leq \mathrm{e}^{-st}\text{,}$$
we have 
$$1-t\leq \mathrm{e}^{-st}\text{.}$$
Then
$$\begin{split}\int_0^1 \frac{\mathrm{e}^{st}}{(q+p\mathrm{e}^{st})^2} (1-t)\mathrm{d}t&\leq \int_0^1 \frac{\mathrm{d}t}{(q+p\mathrm{e}^{st})^2}\\
&\leq\int_0^1 \frac{\mathrm{d}t}{(q+p)^2}\\
&=1\text{.}
\end{split}$$
Therefore
$$\ln(q+p\mathrm{e}^s)\leq ps+pqs^2$$
which is equivalent to
$$q\mathrm{e}^{-ps}+p\mathrm{e}^{qs}\leq\mathrm{e}^{pqs^2}\text{.}$$
