Example 2 Try to find the range of positive numbers $k, k'$ such that the following inequality holds:
$$\frac{\sum(y-z)^{2}}{4 k \sum y z} \leq \sum \frac{x}{y+z}-\frac{3}{2} \leq \frac{\sum(y-z)^{2}}{4 k^{\prime} \sum y z}(x>0, y>0, z>0)$$
The first inequality of (3.2.4) is equivalent to
$$L_{2,1}(x, y, z)=(2 k-1) f_{0,2}^{(5)}+(k-1) f_{0,3}^{(5)}+(6 k-4) f_{1,1}^{(5)} \geq 0$$

The second inequality of (3.2.4) is equivalent to
$$L_{2,2}(x, y, z)=\left(1-2 k^{\prime}\right) f_{0,2}^{(5)}+\left(1-k^{\prime}\right) f_{0,3}^{(5)}+\left(4-6 k^{\prime}\right) f_{1,1}^{(5)} \geq 0$$

Since $y>0$, and
$$\begin{array}{l}
\lim _{z \searrow 0} L_{2.1}(1,1, z)=2(k-1) \geq 0 \\
L_{2.3}(y)=\lim _{z \searrow 0} L_{2,2}(1, y, z)=y\left(1-2 k^{\prime}+k^{\prime} y+k^{\prime} y^{2}+y^{3}-2 k^{\prime} y^{3}\right) \geq 0 \\
\lim _{y \searrow 0} \frac{L_{2.3}(y)}{y}=1-2 k^{\prime} \geq 0
\end{array}$$

Therefore, we get $k \geq 1, k^{\prime} \leq \frac{1}{2}$. Substituting these conclusions into (3.2.5) and (3.2.6), we have
$$L_{2.1}(x, y, z) \geq 0, \quad L_{2.2}(x, y, z) \geq 0$$

Thus, the range of positive numbers $k, k^{\prime}$ that satisfy (3.2.4) is $k \geq 1, 0 \leq k^{\prime} \leq \frac{1}{2}$.