Example 2 Let “ $\sum$ ” denote the cyclic sum. Suppose $a, b, c \in \mathbf{R}_{+}$. Try to find the positive root $x$ of the equation
$$
\sum \sqrt{a b x(a+b+x)}=\sqrt{a b c(a+b+c)}
$$
Solve As shown in Figure 4, construct $\triangle A B C$ with sides $a+b, b+c, a+c$. Draw circles centered at $A, B, C$ with radii $a, b, c$ respectively. Let $\odot O$ be a circle that is externally tangent to the above three circles, with radius $x_{0}$.

Since $S_{\triangle O A B}+S_{\triangle O B C}+S_{\triangle O A C}=S_{\triangle A B C}$, by Heron's formula we have
$$
\sum \sqrt{a b x_{0}\left(a+b+x_{0}\right)}=\sqrt{a b c(a+b+c)} \text {. }
$$

Therefore, $x_{0}$ is the positive root we seek.
Notice that the given conditions are exactly the scenario of four circles being pairwise externally tangent in Descartes' theorem.
From equation (7) we get
$$
x_{0}=\frac{a b c}{a b+b c+c a+2 \sqrt{a b c(a+b+c)}},
$$

which is the positive root we seek.