Example 8 Find the real roots of the equation $\sqrt{x+2 \sqrt{x+2 \sqrt{x+2+\cdots+2 \sqrt{x+2 \sqrt{3 x}}}}}=x$. There are $n$ square roots.
Analysis and Solution: Solving this irrational equation in the usual order from the outside inward is difficult, but the left side of the equation represents the arithmetic root, which implies $x \geqslant 0$, thus each term inside the square roots must be non-negative. This can lead to reverse thinking, starting from the innermost square root, set each term inside the square roots as $y_{1}^{2}, y_{2}^{2}, \cdots, y_{n-1}^{2}, y_{n}^{2}$, then
$$
\begin{array}{l}
3 x=x+2 x=y_{1}^{2} \\
x+2 y_{1}=y_{2}^{2} \\
x+2 y_{2}=y_{3}^{2} \\
\cdots \cdots \\
x+2 y_{n-1}=y_{n}^{2}
\end{array}
$$

At this point, the original equation becomes $y_{n}=x$. To use the above recursive relationship to find $y_{1}$, we can analyze the relationship between $x$ and $y_{1}$: If $x>y_{1}$, then from the comparison of the first two equations, we know $y_{1}>y_{2}$, thus $x>y_{1}>y_{2}>\cdots>y_{n}$, which contradicts $y_{n}=x$. If $x<y_{1}$, it similarly leads to a contradiction. Therefore, it can only be that $y_{1}=x$, hence $3 x=x^{2}$. Solving this yields the two real roots of the original equation as $x_{1}=3, x_{2}=0$.