2 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z \geqslant 6$. Find
$$M=\sum x^{2}+\sum \frac{x}{y^{2}+z+1}$$

the minimum value, where " $\sum$ " denotes the cyclic sum.
2. By the AM-GM inequality, we have $\frac{x^{2}}{14}+\frac{x}{y^{2}+z+1}+\frac{2}{49}\left(y^{2}+z+1\right) \geqslant 3 \sqrt[3]{\frac{x^{3}}{7^{3}}}=\frac{3}{7} x$. Thus, $\frac{1}{14} \sum x^{2}+\sum \frac{x}{y^{2}+z+1}+\frac{2}{49} \sum x^{2}+\frac{2}{49} \sum x+\frac{6}{49} \geqslant \frac{3}{7} \sum x$. Therefore, $\frac{11}{98} \sum x^{2}+\sum \frac{x}{y^{2}+z+1}+\frac{6}{49} \geqslant\left(\frac{3}{7}-\frac{2}{49}\right) \sum x=\frac{19}{49} \sum x$. Also, $\sum x^{2} \geqslant \frac{1}{3}\left(\sum x\right)^{2} \geqslant 12$, so $\sum x^{2}+\sum \frac{x}{y^{2}+z+1}=\frac{87}{98} \sum x^{2}+\frac{11}{98} \sum x^{2}-\frac{6}{49} \geqslant \frac{87}{98} \sum x^{2}+\frac{19}{49} \sum x-\frac{6}{49} \geqslant \frac{87 \times 6+19 \times 6-6}{49}=\frac{90}{7}$. Hence, $M_{\min }=\frac{90}{7}$. At this point, $(x, y, z)=(2,2,2)$