Variation 1. The maximum value of the function $y=3 \sqrt{x-1}+\sqrt{35-7 x}$ is $\qquad$ .
According to the problem, we know that $y=3 \sqrt{x-1}+\sqrt{7} \cdot \sqrt{5-x}>$ 0 and $x \in[1,5]$, so by the Cauchy-Schwarz inequality we get:
$$\begin{array}{c}
y^{2}=[3 \sqrt{x-1}+\sqrt{7} \cdot \sqrt{5-x}]^{2} \leqslant\left[3^{2}+\right. \\
\left.(\sqrt{7})^{2}\right]\left[(\sqrt{x-1})^{2}+(\sqrt{5-x})^{2}\right]=64,
\end{array}$$

Equality holds if and only if $\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{\sqrt{7}}$, which is when $x=\frac{13}{4}$. Therefore, the maximum value of $y=3 \sqrt{x-1}+\sqrt{35-7 x}$ is 8.