Example 2 (1992 "Friendship Cup" International Mathematics Invitational for Grade 9) Let $a, b, c, x, y, z \in R, a^{2}+b^{2}+c^{2}=25, x^{2}+$ $y^{2}+z^{2}=36, a x+b y+c z=30$, find the value of $\frac{a+b+c}{x+y+z}$.
Analysis: From the known equation and the Cauchy-Schwarz inequality, we get
\[ 25 \cdot 36 = \left(a^{2} + b^{2} + c^{2}\right)\left(x^{2} + y^{2} + z^{2}\right) \geqslant (a x + b y + c z)^{2} = 30^{2}. \]
The equality in the above inequality holds if and only if \(\frac{a}{x} = \frac{b}{y} = \frac{c}{z}\). From this, we get
\[ \frac{a+b+c}{x+y+z} = \frac{a}{x} = \frac{b}{y} = \frac{c}{z} = \frac{a x}{x^{2}} = \frac{b y}{y^{2}} = \frac{c z}{z^{2}} = \frac{a x + b y + c z}{x^{2} + y^{2} + z^{2}} = \frac{30}{36} = \frac{5}{6}. \]