Prove $e^{-ax} \le (1-x)^a + \frac{1}{2}ax^2$ for $a > 1$ and $0 \le x \le 1$:

Question

Prove that for all $a > 1$ and $0 \le x \le 1$:

$$ e^{-ax} \le (1-x)^a + \frac{1}{2}ax^2 $$

My very limited start:

$f(x) = (1-x)^a + \frac{1}{2}ax^2 - e^{-ax} \ge 0$

$f'(x) = ax + a e^{-ax} - a(1-x)^{a-1}$

Answer 1

First, note from the well known $e^{t} \geqslant 1+t$ that $$e^{-ax} \geqslant (1-x)^a = (1-x)^{a-1}- x(1-x)^{a-1} \geqslant (1-x)^{a-1}-x$$

Multiply by $-a$ to get $$-a e^{-ax} \leqslant -a(1-x)^{a-1} + ax$$

Integrating this in $[0, x]$ gives $$e^{-ax}\color{red}{-1} \leqslant (1-x)^a\color{red}{-1} + \frac12ax^2$$