3. By the Cauchy inequality, we have $\frac{1}{x^{2}+y+1} \leqslant \frac{1+y+z^{2}}{(x+y+z)^{2}}, \frac{1}{y^{2}+z+1} \leqslant$ $\frac{1+z+x^{2}}{(x+y+z)^{2}}, \frac{1}{z^{2}+x+1} \leqslant \frac{1+x+y^{2}}{(x+y+z)^{2}}$. Therefore, $\frac{1}{x^{2}+y+1}+\frac{1}{y^{2}+z+1}+$ $\frac{1}{z^{2}+x+1} \leqslant \frac{3+x+y+z+x^{2}+y^{2}+z^{2}}{(x+y+z)^{2}}$. Let the right side of the above inequality be $S$. We only need to prove that $S \leqslant 1$. In fact, $S \leqslant 1 \Leftrightarrow 3+x+y+z \leqslant 2(x y+y z+z x)$. Since $x+y+z=$ $x y+y z+z x$, we only need to prove that $x+y+z \geqslant 3$. Also, $x+y+z=x y+y z+z x \leqslant$ $\frac{(x+y+z)^{2}}{3}$, thus, $x+y+z \geqslant 3$. Therefore, the original inequality is proved. The equality holds if and only if $x=y=z=1$.