Using the AM-GM inequality for positive real numbers $x^2/14$, $x/(y^2+z+1)$ and $2(y^2+z+1)/49$ we have
$$\frac{x^2}{14} + \frac{x}{y^2 + z + 1} + \frac{2}{49}(y^2 + z + 1) \ge 3\sqrt[3]{\frac{x^3}{7^3}} = \frac{3}{7}x.$$

We can derive cyclically another two similar inequalities. Adding up of all three derived inequalities we further obtain (all sums are to be considered as sums of three cyclically obtained summands)
$$\frac{1}{14} \sum x^2 - \sum \frac{x}{y^2 + z + 1} + \frac{2}{49} \sum x^2 + \frac{2}{49} \sum x + \frac{6}{49} \ge \frac{3}{7} \sum x.$$
Thus
$$L = \frac{11}{98} \sum x^2 + \sum \frac{x}{y^2 + z + 1} + \frac{6}{49} \ge \left(\frac{3}{7} - \frac{2}{49}\right) \sum x = \frac{19}{49} \sum x.$$

From the Cauchy-Schwarz inequality (or from the AM-QM inequality) it follows $\sum x^2 \ge \frac{1}{3} (\sum x)^2 \ge 12$. Finally we arrive at
$$\begin{aligned} \sum x^2 + \sum \frac{x}{y^2 + z + 1} &= \frac{87}{98} \sum x^2 + L - \frac{6}{49} \ge \frac{87}{98} \sum x^2 + \frac{19}{49} \sum x - \frac{6}{49} \\ &\ge \frac{87 \cdot 6 + 19 \cdot 6 - 6}{49} = \frac{90}{7}. \end{aligned}$$
Conclusion. The smallest value of given expression is therefore $90/7$. (The mentioned value is achieved for $x = y = z = 2$.)