8. Find the minimum value of the function $y=\frac{\sqrt{7 x-3}}{x}, x \in\left[\frac{1}{2}, 3\right]$.

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8. The usual approach is to first convert it into a rational expression of $x$, and then find the extremum. Unfortunately! This can only find the maximum value of $y$. If we start from the given interval $x \in\left[\frac{1}{2}, 3\right]$, we can obtain a clever solution: from $x \in\left[\frac{1}{2}, 3\right]$, we can construct the inequality $\left(x-\frac{1}{2}\right)(x-3) \leqslant 0$, expanding this yields $2 x^{2}-7 x+3 \leqslant 0 \Rightarrow 2 x^{2} \leqslant 7 x-3$,
since $x>0$, then $\sqrt{2} \leqslant \frac{\sqrt{7 x-3}}{x}$,
thus we get $y_{\min }=\sqrt{2}$.