Assume, without any loss, that $a > b$, consider a positive integer $n$, and let $k_n = \lfloor n\sqrt{a} \rfloor - \lfloor n\sqrt{b} \rfloor$. Since $n(\sqrt{a} - \sqrt{b})$ is irrational, it follows that $\{n\sqrt{a}\} \neq \{n\sqrt{b}\}$, so
$$\begin{aligned} 0 < |\{n\sqrt{a}\} - \{n\sqrt{b}\}| &= |n(\sqrt{a} - \sqrt{b}) - k_n| \\ &= \frac{K_n}{(n(\sqrt{a} + \sqrt{b}) - k_n)(n(\sqrt{a} + \sqrt{b}) + k_n)(n(\sqrt{a} - \sqrt{b}) + k_n)}, \end{aligned}$$
where $K_n$ is a positive integer. Finally, notice that $k_n < \lfloor n\sqrt{a} \rfloor + \lfloor n\sqrt{b} \rfloor < n(\sqrt{a} + \sqrt{b})$, to obtain
$$\begin{aligned} |\{n\sqrt{a}\} - \{n\sqrt{b}\}| &> \frac{1}{n(\sqrt{a} + \sqrt{b}) \cdot 2n(\sqrt{a} + \sqrt{b})(n(\sqrt{a} - \sqrt{b}) + n(\sqrt{a} + \sqrt{b}))} \\ &= \frac{1}{4n^3(\sqrt{a} + \sqrt{b})^2\sqrt{a}} = \frac{c}{n^3}, \quad \text{where } c = \frac{1}{4(\sqrt{a} + \sqrt{b})^2\sqrt{a}} > 0. \end{aligned}$$