Showing inequality: $pe^{s(1-p)} + (1-p)e^{-sp} \leq e^{s^2 (3/2) p }$

Question

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!

Answer 1

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. $p\geq \tfrac{1}{4}$.

1. 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 $$\begin{split}\frac{q\mathrm{e}^{\sigma}}{(q+p\mathrm{e}^{\sigma})^2}&\leq \frac{q\mathrm{e}^p}{(q+p\mathrm{e}^p)^2}\\ &\leq \frac{q\mathrm{e}^p}{(q+p)^2}\\ &= (1-p)\mathrm{e}^p\\ &\leq (1-p)(1+\tfrac{p}{1-\tfrac{1}{2}p})\\ &=1- \frac{\tfrac{1}{2}p^2}{1-\tfrac{1}{2}p}\\ &\leq 1\text{.}\end{split}$$
2. 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 $$q\mathrm{e}^{-ps}+p\mathrm{e}^{qs}\leq\mathrm{e}^{pqs^2}\text{.}$$