3. Let $y_{i}=\frac{x_{i+1} x_{i+2}}{x_{i}^{2}}\left(i=1,2, \cdots, n, x_{n+1}=x_{1}, x_{n+2}=x_{2}\right)$, then the original inequality is equivalent to
$$\frac{1}{1+y_{1}}+\frac{1}{1+y_{2}}+\cdots+\frac{1}{1+y_{n}} \leqslant n-1$$

From the definition of $y_{i}$, we know that $y_{1} y_{2} \cdots y_{n}=1$.
If there are two terms in (1) $\leqslant \frac{1}{2}$, since $\frac{1}{1+y_{i}}<1$, the conclusion is already established.
If only $\frac{1}{1+y_{1}} \leqslant \frac{1}{2}$, i.e., $y_{1} \geqslant 1, y_{2} \leqslant 1, y_{3} \leqslant 1, \cdots, y_{n} \leqslant 1$, then
$$\frac{1}{1+y_{1}}+\frac{1}{1+y_{2}} \leqslant \frac{1}{1+y_{1}}+\frac{1}{1+y_{2} y_{3} \cdots y_{n}}=\frac{1}{1+y_{1}}+\frac{1}{1+\frac{1}{y_{1}}}=1$$

Thus, the sum of the first two terms on the left side of (1) $\leqslant 1$, and the remaining $n-2$ terms are all $<1$, so (1) holds.