Prove that $pe^{\alpha q} + qe^{- \alpha p} \le e ^ {\alpha^2/2}$

Question

Prove that,

$$pe^{\alpha q} + qe^{- \alpha p} \le e ^ {\alpha^2/2}$$

where $p$ and $q$ are the probabilities of success and failure in a Bernoulli trial ($0 \le p \le 1, 0 \le q \le 1, p + q = 1$) and $\alpha > 0$.

I tried a bunch of things, but I cannot figure out how to approach it.

Answer 1

let $f(x）=e^{\frac{x^2}{2}}-(pe^{qx}+qe^{-px})$, we will prove $f(x)$ is mono increasing function when $x\ge0$,ie:

$f'(x)>0 \iff xe^{\frac{x^2}{2}}-(pqe^{qx}-pqe^{-px})>0$, it is not trivial,then we prove $f'(x)$ is also mono increasing function, ie:

$f''(x)=e^{\frac{x^2}{2}}+x^2e^{\frac{x^2}{2}}-pq(qe^{qx}+pe^{-px})>0 \iff e^{\frac{x^2}{2}} > pq(qe^{qx}+pe^{-px}) \iff pq \le \frac{1}{4} $

$e^{\frac{x^2}{2}}> \dfrac{1}{4}(qe^{qx}+pe^{-px}) \iff 4e^{\frac{x^2}{2}}>e^{qx}+e^{-px} \iff 4e^{\frac{x^2}{2}}>e^x+1 \iff e^{\frac{x^2}{2}}\ge 1, 3e^{\frac{x^2}{2}}>e^x \iff 3e^{\frac{x^2}{2}-x}>1 \iff 3e^{\frac{(x-1)^2}{2}-\frac{1}{2}}>1 \iff 3>\sqrt{e}$

$f'(0)=0 \implies f'(x)\ge 0 \implies f(0)=0,f(x)\ge0$

QED.