algebra precalculus - Find the smallest positive integer $n$ such that $sqrt{n}-sqrt{n-1}leq frac{1}{100}$

Find the smallest positive integer $n$ such that $\sqrt n-\sqrt{n-1}\le \frac{1}{100}$.

First I multiplied by the conjugate and got

$$\frac 1{\sqrt n + \sqrt{n-1}} \le \frac{1}{100},$$ or $\sqrt n+ \sqrt{n-1} \ge 100$. Now I squared both sides:
$$2n-1+2 \sqrt n \sqrt{n-1}\ge 100.$$
So $$\sqrt n (\sqrt n +\sqrt{n-1})\ge \frac {101}2.$$ However, we know that $\sqrt n+\sqrt{n-1}\ge 100$, so I substituted that:
$$\sqrt n (100) \ge \frac {101}{2}.$$

However, if I solve for $n$ I get around $.25$, and there is no lower positive integer. What did I do wrong? Thanks!

Answer

Let us stop temporarily at

$$\sqrt{n}+\sqrt{n-1}\ge 100.\tag{1}$$
Maybe the algebraic manipulations can stop here.
Note that (1) will certainly be true if $2\sqrt{n-1}\ge 100$, that is, if $n-1 \ge 2500$, But it is conceivable that (1) also holds at $n=2500$. It doesn't.