$$2 \sqrt{x-1} < x$$

I can figure out that since the LHS $\ge$ $0$, $x > 0$ (1) and since the term within the root has to be greater than zero, $x \ge 1$. However, 2 is clearly not acceptable, but there is no way that I can figure how to proceed from here.
$x\ge 1$ is necessary and both terms must be positive so that the inequality is equivalent to this one : $4(x-1)<x^2$ that is $(x-2)^2>0$