Reference for $e^{x-x^2/2a} \le 1+ ax$

Question

I am pretty sure $e^{x-x^2/2a} \le 1+ ax$ for every $x\ge 0$ and $a \ge 1$ but cannot see how to prove it.

Here is an interactive graph.

One idea was the inequality $\log(1+x) \ge x-\frac{x^2}{2}$ that comes from the Taylor series expansion, but that doesn't work because the $a$ is in the wrong place.

This looks like something I've seen before and forgotten. Has anyone else seen it before and remembered?

Answer 1

The claim is false:

For $a = 100$:

Answer 2

In addition to the graphical evidence, here is why you should not expect the inequality (or something similar) to hold:

If we take $x=a$ (the maximizer of the left side), then the inequality reads $$ e^{a/2}\leq 1+a^2. $$ Clearly, this cannot hold for sufficiently large $a$.

Answer 3

Expanding on @MaoWao's comment and @DavidGStork's answer:

$$e^{x-x^2/2a}=e^{-\frac{1}{2a}(x^2-2ax+a^2)}e^{a/2}=e^{-\frac{(x-a)^2}{2a}}e^{a/2}$$ The maximum of the above expression is at $x=a$ and the maximum value is $e^{a/2}$. At the same $x$, the right hand side is $1+a^2$. But we know that the exponential goes to infinity faster than any polynomial, so there is a value $a_0$ for which the left hand side is larger.

Answer 4

Note that the function $f(x) = \ln(1+ ax) - (x-x^2/2a)$ has a local minimum, which has to stay non-negative, in order for the inequality $e^{x-x^2/2a} \le 1+ ax$ to hold for all $x\ge0$.

The critical value of $a$ can be found by requiring that the local minimum of $f(x)$ is zero, i.e. $f'(x)=f(x) = 0$, leading to the following equations

$$\ln(1+ ax) - (x-x^2/2a) = 0$$ $$ax^2-(a^2-1)x-a(a-1)=0$$

The second equation is quadratic, which allows $x$ to be expressed as a function of $a$. Then, plug $x(a)$ into the first equation, which has to be solved numerically and yields $a\approx 8.4$.

Thus, the inequality $e^{x-x^2/2a} \le 1+ ax$ holds for the following values of $a$,

$$1\le a \le 8.4$$