Example 7 In $\triangle A B C$, let $B C=a, A C=b, A B=c$. If $\angle A+\angle C=2 \angle B$. Find the largest positive integer $n$, such that $a^{n}+c^{n} \leqslant 2 b^{n}$ holds for any such triangle.
Analysis We first estimate the range of $n$, then proceed to prove it. Here, we use a special triangle for estimation.
Solution From $\angle A+\angle C=2 \angle B$, it is easy to know: $\angle B=60^{\circ}$, and $\angle A+\angle C=120^{\circ}$. By the Law of Sines, $a^{n}+c^{n} \leqslant 2 b^{n}$ is equivalent to $\sin ^{n} A+\sin ^{n} C \leqslant 2 \sin ^{n} B$. Clearly, $\angle A=90^{\circ}, \angle C=30^{\circ}$ is a triangle that satisfies the condition.
So $1^{n}+\left(\frac{1}{2}\right)^{n} \leqslant 2\left(\frac{\sqrt{3}}{2}\right)^{n}$ holds.
When $n \geqslant 5$, $2\left(\frac{\sqrt{3}}{2}\right)^{n}<1$, so $n \leqslant 4$.
Next, we prove that when $n=4$, $a^{n}+c^{n} \leqslant 2 b^{n}$ always holds.
By the Law of Cosines: $b^{2}=a^{2}+c^{2}-2 a c \cos B=a^{2}+c^{2}-a c$, then $2 b^{4}-a^{4}-c^{4}=2\left(a^{2}+c^{2}-a c\right)^{2}$ $-a^{4}-c^{4}=a^{4}+c^{4}+4 a^{2} c^{2}+2 a^{2} c^{2}-4 a^{3} c-4 a c^{3}=\left(a^{2}+c^{2}-a c\right)^{2} \geqslant 0$, so when $n=4$, $2 b^{4} \geqslant$ $a^{4}+c^{4}$ always holds. In summary, the maximum value of $n$ is 4.