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).