4.3.5 ** If $a, b, c, x, y$ and $z$ are all real numbers, and
$$
a^{2}+b^{2}+c^{2}=25, x^{2}+y^{2}+z^{2}=36, a x+b y+c z=30 \text{. }
$$

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

the equality holds. From the condition for equality in the Cauchy-Schwarz inequality, we know that $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k$. Thus, $k^{2}\left(x^{2}+y^{2}+z^{2}\right)=25$, so $k=+\frac{5}{6}$ (the negative value is discarded). Therefore, using the theorem of the ratio of sums, we have
$$
\frac{a+b+c}{x+y+z}=k=\frac{5}{6} .
$$