# Showing inequality: $pe^{x(1-p)}+(1-p)e^{-xp} \leq e^{x^2(3/4)p}$ for $0 \leq p \leq 1/2, 0 \leq x \leq 1$?

How can I show that $$pe^{x(1-p)}+(1-p)e^{-xp} \leq e^{x^2(3/4)p}$$ for $0 \leq p \leq 1/2, 0 \leq x \leq 1$?

I've been stuck on this for a long time; I tried expanding out the taylor series on either side, and I tried using convexity, but neither method seems to help. I was able to get the inequality down to $\leq e^{x^2(1-p)p}$ on the right side using the method here but I couldn't see a way to further tighten the upper bound.

• This problem's a bit different, it has different bounds (e.g. $0 \leq p \leq 1/2, 0 \leq x \leq 1$ rather than $0 \leq p \leq 1/2, 0 \leq x \leq p$. Commented Mar 28, 2018 at 16:37
• Where did you come across this problem? Another interesting upper bound is this: math.stackexchange.com/questions/1978987 I wonder if there are information theory versions of the bounds. Commented Aug 19, 2018 at 12:12

We show that if $0\leq x<3$ and $0\leq p \leq 1$ then $$(1-p)\mathrm{e}^{-px}+p\mathrm{e}^{(1-p)x}\leq \mathrm{e}^{\tfrac{1}{2}px^2(1+R(x))}$$ where $$R(x)=\frac{x}{3-x}\text{.}$$

From the integral form of the remainder for Taylor's theorem, we have $$\ln((1-p)+p\mathrm{e}^{x})=px +px^2 \int_0^1 \frac{(1-p)\mathrm{e}^{xt}}{((1-p)+p\mathrm{e}^{xt})^2} (1-t)\mathrm{d}t\text{.}$$ For fixed $t$ and $x\geq 0$, we have

$$\frac{\mathrm{d}}{\mathrm{d}p}\frac{(1-p)\mathrm{e}^{xt}}{(1-p+p\mathrm{e}^{xt})^2}=-\frac{\mathrm{e}^{xt}}{(1-p+p\mathrm{e}^{xt})^2}-\frac{2(1-p)\mathrm{e}^{xt}(\mathrm{e}^{xt}-1)}{(1-p+p\mathrm{e}^{xt})^2}\leq 0\text{.}$$ Therefore the integrand is antitone in $p$ for fixed $x,t$, whence

$$\begin{split}\int_0^1 \frac{(1-p)\mathrm{e}^{xt}}{((1-p)+p\mathrm{e}^{xt})^2} (1-t)\mathrm{d}t&\leq\int_0^1\mathrm{e}^{xt}(1-t)\mathrm{d} t\\ &=\frac{\mathrm{e}^x-1-x}{x^2} \end{split}$$

But if $0\leq x<3$, then $$\begin{split} \mathrm{e}^x&\leq 1+x+ \frac{\tfrac{x^2}{2}}{1-\tfrac{x}{3}} \end{split}$$ so that $$\int_0^1 \frac{(1-p)\mathrm{e}^{xt}}{((1-p)+p\mathrm{e}^{xt})^2} (1-t)\mathrm{d}t\leq \frac{1}{2}+\frac{x}{2(3-x)}\text{.}$$

Thus $$\ln((1-p)+p\mathrm{e}^{x})\leq px +\tfrac{1}{2}px^2\left(1+R(x)\right)$$ where $$R(x)=\frac{x}{3-x}\text{.}$$ The desired result follows by exponentiation.

Note \begin{align*} p e^{x(1-p)}+(1-p)e^{-xp} = e^{-xp}(1-p+p e^{x}) \le e^{(e^x-1-x)p}. \end{align*} (Using the standard $1+x\le e^x$ for $x\in\mathbb R$.)

Now $$(e^x-1-x) =\sum_{k\ge2}x^k/k! \le x^2\sum_{k\ge2}1/k!=x^2(e-2)\le0.72 x^2\le \frac{3}{4}x^2.$$ (where we used $0\le x\le 1$.)

Putting the two together gives your inequality.