This is a follow-up to this question. I came across a tighter potential bound and checked it numerically for $0\le x\le 5$. I think it holds for all positive $x$, can anyone see a proof?
$$1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$$
Note: using analysis for previous question you can show that $1-\exp(-k x^2)$ is an upper bound on $\text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/\pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $\text{erf}(x)^2$ very closely.
Dashed graph below is $\text{erf}(x)^2$, red is $1-\exp(-k x^2)$ for $k=4/\pi$, other two graphs are for $k=1$ and $k=2$
Best Answer
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).