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

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.

Any help pointing to a useful link would also be greatly appreciated. I am having difficulties with solving rational inequalities, specially ones that involve a quadratic expression with roots/modulus, and I don't have a text book with me as of now. Any online link helping me out wouuld be just great! Thanks !!

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$x\ge 1$ is necessary and both terms must be positive so that the inequality is equivalent to this one : $4(x-1)<x^2$ that is $(x-2)^2>0$