Let $BC = a$, $CA = b$, $AB = c$, $PA' = \lambda$, $PB' = u$, $PC' = v$. In the theorem, let
$$\begin{array}{c} x' = -a^2 + b^2 + c^2 \\ y' = a^2 - b^2 + c^2 \\ z' = a^2 + b^2 - c^2 \\ x = \frac{\lambda}{a}, y = \frac{u}{b}, z = \frac{v}{c} \end{array}$$

Then the left side of equation $\left(1'\right)$ is
$$\frac{(y' + z')}{2} x + \frac{(z' + x')}{2} y + \frac{(x' + y')}{2} z = \lambda a + u b + v c = 2 \Delta$$

At this time, the right side of equation (1) is
$$\sqrt{y' z' + z' x' + x' y'} \cdot \sqrt{y z + z x + x y} = 4 \Delta \cdot \sqrt{\frac{u v}{b c} + \frac{v \lambda}{c a} + \frac{\lambda u}{a b}} = 4 \Delta \cdot \sqrt{\frac{\frac{1}{2} u v \sin A}{\Delta} + \frac{\frac{1}{2} v \lambda \sin B}{\Delta} + \frac{\frac{1}{2} \lambda u \sin C}{\Delta}}$$
Note that $\frac{1}{2} u v \sin A = \frac{1}{2} u v \sin \angle C' P B' = S_{\Delta B' C' P}$, etc.
$$\text{The above expression} = 4 \Delta \cdot \sqrt{\frac{\Delta'}{\Delta}} = 4 \sqrt{\Delta \Delta'}$$

Therefore, we have
$$\begin{aligned} 2 \Delta & \geqslant 4 \sqrt{\Delta \Delta'} \\ \Delta' & \leqslant \frac{1}{4} \Delta \end{aligned}$$