Proof that $\text{erfc}(x)\leqslant e^{-x^2}$

Question

I'm looking for a proof that $\text{erfc}(x)\leqslant e^{-x^2}$ without invoking any probability theory.

Some sources say this inequality is a result of the "Chernoff bound," which seems to come from probability theory. But in principle this inequality is about real-valued functions and should not require any reference to probability theory.

Is there a simple way to prove this?

Answer 1

For $x \ge 0$ one can substitute $t = x+s$ in the integral: $$ \operatorname{erfc}(x) = \frac{2}{\sqrt \pi} \int_x^\infty e^{-t^2} \, dt = \frac{2}{\sqrt \pi} \int_0^\infty e^{-(x+s)^2} \, ds \\ = \frac{2}{\sqrt \pi} e^{-x^2 }\int_0^\infty e^{-2xs}e^{-s^2} \, ds \le \frac{2}{\sqrt \pi} e^{-x^2 }\int_0^\infty e^{-s^2} \, ds \\ = e^{-x^2 }\operatorname{erfc}(0) = e^{-x^2 } $$

For $x< 0$ the estimate does not hold because then $e^{-x^2} < 1 < \operatorname{erfc}(x)$.

Answer 2

For $x \ge \dfrac{1}{\sqrt{\pi}}$, we have $\dfrac{t}{x} \ge 1$ for all $t \in [x,\infty)$, and thus $$\text{erfc}(x) = \dfrac{2}{\sqrt{\pi}}\int_{x}^{\infty}e^{-t^2}\,dt \le \dfrac{2}{\sqrt{\pi}}\int_{x}^{\infty}\dfrac{t}{x}e^{-t^2}\,dt = \dfrac{1}{x\sqrt{\pi}}e^{-x^2} \le e^{-x^2}.$$

Also, for $0 \le x \le \dfrac{1}{\sqrt{\pi}}$, we have $t\sqrt{\pi} \le 1$ for all $t \in [0,x]$, and thus

$$1-\text{erfc}(x) = \dfrac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}\,dt \ge \dfrac{2}{\sqrt{\pi}}\int_{0}^{x}t\sqrt{\pi}e^{-t^2}\,dt = 1-e^{-x^2},$$ i.e. $\text{erfc}(x) \le e^{-x^2}$.