Prove: Since
$$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cdot \cos B \cdot \cos C=1$$

Therefore,
$$(2 \cos A)^{2}+(2 \cos B)^{2}+(2 \cos C)^{2}+(2 \cos A) \cdot(2 \cos B) \cdot(2 \cos C)=4$$

Thus, the triplet $(2 \cos A, 2 \cos B, 2 \cos C)$ is a solution to the equation $x^{2}+y^{2}+z^{2}+x y z=4$.

Conversely, it is easy to see that $x, y, z \in(0,2)$, so there exist $\angle A, \angle B \in\left(0, \frac{\pi}{2}\right)$ such that $x=2 \cos A, y=2 \cos B$.

Solving the equation $x^{2}+y^{2}+z^{2}+x y z=4$ for $z$, we get
$$z=-2 \cos (A+B)$$

Thus, let $\angle C=\pi-(\angle A+\angle B)$, then the real solutions to the equation
$x^{2}+y^{2}+z^{2}+x y z=4$ are
$$(x, y, z)=(2 \cos A, 2 \cos B, 2 \cos C)$$

Analysis: The structure and degree relationship of the conditions and the lemma are similar, so by making a variable substitution, the structure and form become more apparent, thus directly applying the J.Wolstenholme-embedded inequality to solve the problem.
Proof: Let: $u=x^{2}, v=y^{2}, w=z^{2}$.
Then the equation $u+v+w+\sqrt{u v w}=4$ is equivalent to the equation
$$x^{2}+y^{2}+z^{2}+x y z=4$$

According to the lemma, if and only if there exists an acute triangle $\triangle A B C$ such that
$$x=2 \cos A, y=2 \cos B, z=2 \cos C$$

And the inequality to be proved, $\sqrt{\frac{v w}{u}}+\sqrt{\frac{u w}{v}}+\sqrt{\frac{u v}{w}} \geq u+v+w$, is equivalent to
$$\frac{y z}{x}+\frac{z x}{y}+\frac{x y}{z} \geq x^{2}+y^{2}+z^{2}$$

Without loss of generality, let $x=2 \cos A, y=2 \cos B, z=2 \cos C$.
Then we only need to prove the inequality
$$\begin{array}{l} \frac{2 \cos B \cdot \cos C}{\cos A}+\frac{2 \cos C \cdot \cos A}{\cos B}+\frac{2 \cos A \cdot \cos B}{\cos C} \\ \geq 4 \cos ^{2} A+4 \cos ^{2} B+4 \cos ^{2} C \end{array}$$

By the J.Wolstenholme-embedded inequality, we have
$$\begin{array}{l} \frac{2 \cos B \cdot \cos C}{\cos A}+\frac{2 \cos C \cdot \cos A}{\cos B}+\frac{2 \cos A \cdot \cos B}{\cos C} \\ \geq 2 \sqrt{\frac{2 \cos C \cdot \cos A}{\cos B}} \cdot \sqrt{\frac{2 \cos A \cdot \cos B}{\cos C}} \cdot \cos A \\ +2 \sqrt{\frac{2 \cos A \cdot \cos B}{\cos C}} \cdot \sqrt{\frac{2 \cos B \cdot \cos C}{\cos A}} \cdot \cos B \\ +2 \sqrt{\frac{2 \cos B \cdot \cos C}{\cos A}} \cdot \sqrt{\frac{2 \cos C \cdot \cos A}{\cos B}} \cdot \cos C \\ =4 \cos ^{2} A+4 \cos ^{2} B+4 \cos ^{2} C \end{array}$$

Thus, the original inequality is proved!