Example 9 The range of the function $y=\sin ^{12} x+\cos ^{12} x$ is $\qquad$
Solution 1: By the AM-GM inequality, we know that
$$
\left(\sin ^{4} x\right)^{3}+\left(\frac{1}{4}\right)^{3}+\left(\frac{1}{4}\right)^{3} \geqslant 3 \sin ^{4} x \cdot\left(\frac{1}{4}\right)^{2},
$$
$$
\left(\cos ^{4} x\right)^{3}+\left(\frac{1}{4}\right)^{3}+\left(\frac{1}{4}\right)^{3} \geqslant 3 \cos ^{4} x \cdot\left(\frac{1}{4}\right)^{2}.
$$

Adding the two inequalities, we get $\sin ^{12} x+\cos ^{12} x \geqslant \frac{3}{16}\left(\sin ^{4} x+\cos ^{4} x\right)-\frac{1}{16}$
$$
=\frac{3}{16}\left(1-2 \sin ^{2} x \cos ^{2} x\right)-\frac{1}{16}=-\frac{3}{32} \sin ^{2} 2 x+\frac{1}{8} \geqslant \frac{1}{32}.
$$

When $x=\frac{\pi}{4}$, the above inequalities hold with equality.
Thus, $y_{\text {min }}=\frac{1}{32}$.
Since $\sin x, \cos x \in[-1,1]$,
we have
$$
y=\sin ^{12} x+\cos ^{12} x \leqslant \sin ^{2} x+\cos ^{2} x=1.
$$

Therefore, $y_{\max }=1$.
Hence, the range of $y$ is $\left[\frac{1}{32}, 1\right]$.
Solution 2: By the Cauchy-Schwarz inequality, we know that
$$
\begin{array}{l}
\sin ^{12} x+\cos ^{12} x=\frac{1}{2}(1+1)\left(\sin ^{12} x+\cos ^{12} x\right) \geqslant \frac{1}{2}\left(\sin ^{6} x+\cos ^{6} x\right)^{2} \\
=\frac{1}{2}\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right)^{2}=\frac{1}{2}\left(1-\frac{3}{4} \sin ^{2} 2 x\right)^{2} \geqslant \frac{1}{32}.
\end{array}
$$

Since $\sin x, \cos x \in[-1,1]$, we have $\sin ^{12} x+\cos ^{12} x \leqslant \sin ^{2} x+\cos ^{2} x=1$,
Thus, the range of $y$ is $\left[\frac{1}{32}, 1\right]$.
Solution 3: Since $\sin ^{12} x+\cos ^{12} x=\left(\sin ^{12} x+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\right)+\left(\cos ^{12} x+\frac{1}{64}+\frac{1}{64}+\right.$
$$
\begin{array}{l}
\left.\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\right)-\frac{10}{64} \geqslant 6 \sqrt[6]{\sin ^{12} x\left(\frac{1}{64}\right)^{5}}+6 \sqrt[6]{\cos ^{12} x\left(\frac{1}{64}\right)^{5}}-\frac{5}{32}=6 \cdot \frac{1}{2^{5}}\left(\sin ^{2} x+\cos ^{2} x\right)-\frac{5}{32}= \\
\frac{6}{32}-\frac{5}{32}=\frac{1}{32}, \text { and } \sin ^{12} x+\cos ^{12} x \leqslant\left(\sin ^{2} x+\cos ^{2} x\right)^{6}=1,
\end{array}
$$

Therefore, $y \in\left[\frac{1}{32}, 1\right]$.