By rewriting $f(x)$, we have
$$f(x) = \frac{\sqrt{2}\sin\left(\pi x - \frac{\pi}{4}\right) + 2}{\sqrt{x}}$$
for $\frac{1}{4} \le x \le \frac{5}{4}$. Define $g(x) = \sqrt{2}\sin(\pi x - \frac{\pi}{4})$, where $\frac{1}{4} \le x \le \frac{5}{4}$. Then $g(x) \ge 0$, and $g(x)$ is monotone increasing on $[\frac{1}{4}, \frac{3}{4}]$, and monotone decreasing on $[\frac{3}{4}, \frac{5}{4}]$.

Further, the graph of $y = g(x)$ is symmetric about $x = \frac{3}{4}$, i.e. for any $x_1 \in [\frac{1}{4}, \frac{3}{4}]$ there exists $x_2 \in [\frac{3}{4}, \frac{5}{4}]$ such that $g(x_2) = g(x_1)$. Then
$$f(x_1) = \frac{g(x_1) + 2}{\sqrt{x_1}} = \frac{g(x_2) + 2}{\sqrt{x_1}} \ge \frac{g(x_2) + 2}{\sqrt{x_2}} = f(x_2).$$
On the other hand, $f(x)$ is monotone decreasing on $[\frac{3}{4}, \frac{5}{4}]$. Therefore $f(x) \ge f(\frac{5}{4}) = \frac{4\sqrt{5}}{5}$. That means the minimum value of $f(x)$ on $[\frac{1}{4}, \frac{5}{4}]$ is $\frac{4\sqrt{5}}{5}$.