Proving $1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$

As before we consider

$$\text{erf}(x)^2={4\over \pi}\int_0^x\int_0^x \exp{-(s^2+t^2)}\ ds \ dt\,.$$

Now compare this with the same over the area which is given by the quarter of a circle of radius $\displaystyle \frac{2x}{\sqrt{\pi}}$. The area of this is same as the area of the square of side $x$.

Since $\displaystyle e^{-(s^2 + t^2)}$ decreases as $\displaystyle s^2 + t^2$ increases, we are done!

The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $\displaystyle s^2 + t^2$ is higher in that region).

This is one of the many results that keep being periodically rediscovered. The bound, together with a less accurate lower bound of the same form, was given in a report by J.T. Chu (1954; On bounds for the normal integral). It was noted in the report that the upper bound had previously been obtained independently by G. Polya and J.D. Williams. Both used the nice geometrical argument that has been here expounded.