5. Three positive numbers $x, y, z$ satisfy the system of inequalities:
$$
\left\{\begin{array}{l}
\frac{11}{6} z < x + y < 2 \\
\frac{3}{2} x < y + z < \frac{5}{3} x \\
\frac{5}{2} y < x + z < \frac{11}{1} y .
\end{array}\right.
$$

Then the relationship in size of $x, y, z$ is ().
(A) $x < y < z$
(B) $y < x < z$
(C) $z < x < y$
(D) The size relationship cannot be determined
5. B

From $\left\{\begin{array}{l}\frac{11}{6}\frac{5}{2}>2>\frac{x+y}{z}>\frac{11}{6}>\frac{5}{3}>\frac{y+z}{x}$.
Thus, $z(x+z)>y(x+y) \Rightarrow z>y$. (Otherwise, the inequality does not hold)
$x(x+y)>z(y+z) \Rightarrow x>z$. (For the same reason)
Therefore, $y<z<x$.