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. 
 A: 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.
A: 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.
