Czech-Slovak-Polish Match

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 $$\begin{array}{rl}\sum x^2+\sum \frac{x}{y^2+z+1}& =\frac{87}{98}\sum x^2+L-\frac{6}{49}\ge \frac{87}{98}\sum x^2+\frac{19}{49}\sum x-\frac{6}{49}\\ & \ge \frac{87\cdot 6+19\cdot 6-6}{49}=\frac{90}{7}.\end{array}$$ Conclusion. The smallest value of given expression is therefore $90/7$. (The mentioned value is achieved for $x=y=z=2$.)