China Mathematical Competition

First, from $$\{\begin{array}{l}x+2y>0,\\ x-2y>0,\\ (x+2y)(x-2y)=4,\end{array}$$ we obtain $$\{\begin{array}{l}x>2|y|\ge 0,\\ x^2-4y^2=4.\end{array}$$

By the symmetry, there is no loss of generality in considering only the case when $y\ge 0$. In view of $x>0$, we need to find the minimum value of $x-y$ only. Setting $u=x-y$, and substituting it into $x^2-4y^2=4$, we obtain $$3y^2-2uy+(4-u^2)=0.(\ast )$$ Equation (*) with respect to $y$ has real solutions. So we have $$\Delta =4u^2-12(4-u^2)\ge 0.$$ Thereby $$u\ge \sqrt{3}.$$

In addition, when $x=\frac{4}{3}\sqrt{3}$ and $y=\frac{\sqrt{3}}{3}$, we have $u=\sqrt{3}$. Therefore, the minimum value of $|x|-|y|$ is $\sqrt{3}$.