Different methods for finding the minimum of $|x-2y|$ when $x^2+1=2y^2$. For $x, y \in \Bbb R$, $x^2 + 1 = 2y^2$, find the minimum of $|x - 2y|$.
At a glance I found that the point $(x, y)$ lies on a hyperbola and $|x - 2y|$ is just the distance between the point and the line $x - 2y = 0$ times $\sqrt 5$. Now this problem can be solved by using derivatives.
My question is: are there any other solutions? I want to find some other beautiful solutions to this interesting problem. All methods are accepted as long as they are rigorous.

Calculations give us
$$(x-2y)^2=2(x-y)^2+1 \ge 1$$
Thus we have $$|x-2y| \ge 1$$ With the equality holding when $x=y=\pm 1$. 
