Example 4 Solve the equation $\sqrt{12-\frac{12}{x^{2}}}+\sqrt{x^{2}-\frac{12}{x^{2}}}=x^{2}$.
Solve the original equation, which can be rewritten as $\frac{\sqrt{12}}{x^{2}} \cdot \sqrt{1-\frac{1}{x^{2}}}+\frac{1}{|x|}$. $\sqrt{1-\frac{12}{x^{4}}}=1$, and by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(\frac{\sqrt{12}}{x^{2}} \cdot \sqrt{1-\frac{1}{x^{2}}}+\frac{1}{|x|} \cdot \sqrt{1-\frac{12}{x^{4}}}\right)^{2} \\
\leqslant & {\left[\left(\frac{\sqrt{12}}{x^{2}}\right)^{2}+\left(\sqrt{1-\frac{12}{x^{4}}}\right)^{2}\right] } \\
& \cdot\left[\left(\sqrt{1-\frac{1}{x^{2}}}\right)^{2}+\left(\frac{1}{|x|}\right)^{2}\right]=1
\end{aligned}$$

The equality holds if there exists a constant $k$ such that
$$\left\{\begin{array}{l}
\frac{\sqrt{12}}{x^{2}}=k \cdot \sqrt{1-\frac{1}{x^{2}}} \\
\sqrt{1-\frac{12}{x^{4}}}=k \cdot \frac{1}{|x|}
\end{array}\right.$$

Thus, the original equation is equivalent to (4). Squaring both sides of (4) and adding them yields $k^{2}=1$, hence $\frac{12}{x^{4}}=1-\frac{1}{x^{2}}$. From this, it is easy to solve for $x= \pm 2$. Upon verification, $x= \pm 2$ are indeed the roots of the original equation.