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. 

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.
