China Mathematical Competition

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 $\frac{4\sqrt{5}}{5}$.