Example 11 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 the Cauchy-Schwarz inequality, we have
$$25 \times 36=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geqslant(a x+b y+c z)^{2}=30^{2} .$$

From the equality condition of the above inequality, we get \(a=k x, b=k y, c=k z\).
Thus, \(25=a^{2}+b^{2}+c^{2}=k^{2} x^{2}+k^{2} y^{2}+k^{2} z^{2}\)
$$=k^{2}\left(x^{2}+y^{2}+z^{2}\right)=36 k^{2},$$
\(\therefore k=\frac{5}{6}\) (note that \(a x+b y+c z=30, k=-\frac{5}{6}\) should be discarded).
$$\text { Therefore, } \frac{a+b+c}{x+y+z}=k=\frac{5}{6} \text {. }$$