For real numbers $a, b, c, x, y, z$,

\[
\begin{aligned}
& a^{2}+b^{2}+c^{2}=25 \\
& x^{2}+y^{2}+z^{2}=36 \\
& a x+b y+c z=30
\end{aligned}
\]

What is the value of the ratio $\frac{a+b+c}{x+y+z}$?
According to the well-known Cauchy-Schwarz-Bunyakovszkij inequality,

$$
\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geqq(a x+b y+c z)^{2}
$$

for any real numbers \(a, b, c, x, y, z\), and in (1) equality holds if and only if there exists a real number \(\lambda\) such that

$$
a=\lambda x, \quad b=\lambda y \quad \text { and } \quad c=\lambda z \text {. }
$$

Notice that according to the conditions of the problem, equality holds in (1): the value of \(a x+b y+c z\) is 30, which is precisely the geometric mean of the given values of \(a^{2}+b^{2}+c^{2}\) and \(x^{2}+y^{2}+z^{2}\), 25 and 36, respectively. Thus, using the condition for equality,

$$
a^{2}+b^{2}+c^{2}=\lambda^{2}\left(x^{2}+y^{2}+z^{2}\right)
$$

from which

$$
\lambda^{2}=25 / 36, \quad \text { so } \quad|\lambda|=\frac{5}{6}
$$

We also know that

$$
30=a x+b y+c z=\lambda\left(x^{2}+y^{2}+z^{2}\right)
$$

therefore \(\lambda\) is positive, i.e., \(\lambda=5 / 6\)

The value of the ratio in question is then

$$
\frac{a+b+c}{x+y+z}=\frac{\lambda(x+y+z)}{x+y+z}=\lambda
$$

so the desired value is \(5 / 6\).

Remark. Clearly, there exist real numbers that satisfy the given conditions, for example \(a=5, b=c=0 ; x=6 ; y=z=0\).