Example 5 Let $a, b, c, x, y, z$ be real numbers, and
$$a^{2}+b^{2}+c^{2}=25, x^{2}+y^{2}+z^{2}=36, a x+b y+c z=30 .$$

Find the value of $\frac{a+b+c}{x+y+z}$.
By Cauchy-Schwarz inequality, we have
$$25 \times 36=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geqslant(a x+b y+c z)^{2}=30^{2} .$$

For the equality to hold, we get
$$\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k$$

Thus, \( k^{2}\left(x^{2}+y^{2}+z^{2}\right)=25 \), so \( k= \pm \frac{5}{6} \) (the negative value is discarded). Therefore,
$$\frac{a+b+c}{x+y+z}=k=\frac{5}{6} .$$