Example 4 When point $P$ moves along the line $y=2 x+6$, and point $Q$ moves on the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{4}=1$, the minimum length of segment $P Q$ is $\qquad$
Solution: Let $Q\left(x_{0}, y_{0}\right)$ be the point whose distance to the line $y=2 x+6$ is $d$. Since point $Q$ lies on the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{4}=1$, we have $\frac{x_{0}{ }^{2}}{6}+\frac{y_{0}^{2}}{4}=1$. Also, $|P Q| \geqslant d=\frac{\left|2 x_{0}-y_{0}+6\right|}{\sqrt{5}}$. By the Cauchy-Schwarz inequality, we get $\left(2 \sqrt{6} \cdot \frac{x_{0}}{\sqrt{6}}+2 \cdot \frac{-y_{0}}{2}\right)^{2} \leqslant$ (24 +4) $\left(\frac{x_{0}{ }^{2}}{6}+\frac{y_{0}{ }^{2}}{4}\right)$, thus $-2 \sqrt{7} \leqslant 2 x_{0}-y_{0} \leqslant 2 \sqrt{7}$. Therefore, $2 x_{0}-y_{0}+6>0$, so $|P Q| \geqslant d=$ $\frac{\left|2 x_{0}-y_{0}+6\right|}{\sqrt{5}}=\frac{2 x_{0}-y_{0}+6}{\sqrt{5}} \geqslant \frac{6-2 \sqrt{7}}{\sqrt{5}}=$ $\frac{6 \sqrt{5}-2 \sqrt{35}}{5}$. Therefore, the minimum length of segment $P Q$ is $\frac{6 \sqrt{5}-2 \sqrt{35}}{5}$.