Estimate $e^{-x}$ in formular

Question

Do anyone has a hint how to proof by estimate $ e^{-ax} \leq (1-x)^{a} + \frac{1}{2}ax^2 $ for $x,a \in \mathbb{R}$ and $0 \leq x \leq 1$ and $a > 1$.

Simple starting with the definition of $e$ leads me to (if I am not wrong):

$e^{-xa} = \sum_{n \geq 0} \frac{(-1)^n}{n!}(ax)^n = 1 - ax + \frac{(ax)^2}{2}-\frac{(ax)^3}{6}+...$.

But the estimation needs a tiny trick, my ideas with estimate are too rough. I just began with thinking that $-a \leq -ax \leq 0$ and then trying to extrude it to all the other terms, but I don't know how to go further.

Otherwise it is easy to show that $e^{-x} \geq 1-x$ but this is the opposite beginning of what I want so I stop with that, too.

Thank you in advance. If you just have an input this is enough, I may want to solve it on my own, if there is an idea how to start.

Answer 1

Let $g(x)$ be given by

$$g(x)=(1-x)^a+\frac12ax^2-e^{-ax}$$

for $x\in[0,1]$ and $a>1$.

Applyng Bernoulli's Inequality, we see that for $x\in[0,1]$ and $a\ge1$

$$g(x)\ge 1-ax+\frac12ax^2-e^{-ax}$$

Let $h(x)= 1-ax+\frac12ax^2-e^{-ax}$. Note that $h(0)=0$ and for $a\ge 1$

$$h'(x)=a(e^{-ax}-(1-x))\ge a(e^{-ax}-(1-ax))\ge 0$$

From the mean value theorem, $h(x)\ge 0$ and therefore for $x\in[0,1]$ and $a\ge 1$

$$h(x)\ge 0\implies g(x)\ge 0\implies e^{-ax}\le (1-x)^a+\frac12ax^2$$

as was to be shown!

Answer 2

If we set $$ f_a(x) = (1-x)^a e^{ax}+\frac{a}{2}x^2 e^{ax} $$ we have $$ f_a'(x) = a e^{a x} x \left(1-(1-x)^{a-1}+\frac{a}{2}x\right)$$ from which it is trivial that $f_a(x)$ is increasing on $(0,1)$. Since $f_a(0)=1$ it follows that $$ \forall x\in[0,1],\qquad (1-x)^a+\frac{a}{2}x^2 \geq e^{-ax}.$$