Kinematics is Easy, Isn't it?

Let's define $\vec{x}_{1}(t)$ and $\vec{x}_{2}(t)$ as left and right vector position, respectively. If we set the origin of the coordinate system on the first particle, we have:

$\vec{x}_{1}(t)=\frac{vt}{\sqrt{2}} \hat{i} + \frac{vt}{\sqrt{2}} \hat{j}$

$\vec{x}_{2}(t)= x_{0}\hat{i}+vt\hat{j}$

The modulus of the difference vector between them is the distance between the particles at any moment:

$d(t)=\|\vec{x}_{2}(t)-\vec{x}_{1}(t)\|=\sqrt{(\frac{vt}{\sqrt{2}}-x_{0})^2+(\frac{vt}{\sqrt{2}}-vt)^2}$

$=\sqrt{\frac{v^2t^2}{2}-\sqrt{2}vtx_{0}+x_{0}^2+v^2t^2(\frac{1}{2}-\sqrt{2}+1)}$

$=\sqrt{v^2(2-\sqrt{2})t^2-\sqrt{2}x_{0}vt+x_{0}^2}$

As square root function is inscreasing, its minimum value is reached when its argument is minimum:

$argmin(d(t))=\frac{\sqrt{2}vx_{0}}{2(2-\sqrt{2})v^2}=\frac{x_{0}}{2(\sqrt{2}-1)v}$

And finally:

$min(d(t))=\sqrt{v^2(2-\sqrt{2})\frac{x_{0}^2}{4(3-2\sqrt{2})v^2}-\sqrt{2}x_{0}v\frac{x_{0}}{2(\sqrt{2}-1)v}+x_{0}^2}$

$=x_{0}\sqrt{\frac{2+\sqrt{2}}{4}-\frac{2+\sqrt{2}}{2}+1}$

$=\frac{x_{0}}{2}\sqrt{2-\sqrt{2}}$