Solution:
First we factor $x^{5}+x+1$ as follows:
$$\begin{aligned} x^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\left(x^{3}-1\right)+x^{2}+x+1=x^{2}(x-1)\left(x^{2}+x+1\right)+x^{2}+x+1 \\ & =\left(x^{2}+x+1\right)\left(x^{2}(x-1)+1\right)=\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right) \end{aligned}$$
Using the AM-GM inequality, we have
$$\sqrt{x^{5}+x+1}=\sqrt{\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)} \leq \frac{x^{2}+x+1+x^{3}-x^{2}+1}{2}=\frac{x^{3}+x+2}{2}$$
and since
$x^{3}+x+2=x^{3}+1+x+1=(x+1)\left(x^{2}-x+1\right)+x+1=(x+1)\left(x^{2}-x+1+1\right)=(x+1)\left(x^{2}-x+2\right)$,
then
$$\sqrt{x^{5}+x+1} \leq \frac{(x+1)\left(x^{2}-x+2\right)}{2}$$
Using $x^{2}-x+2=\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}>0$, we obtain $\frac{x+1}{\sqrt{x^{5}+x+1}} \geq \frac{2}{x^{2}-x+2}$ Applying the Cauchy-Schwarz inequality and the given condition, we get
$$\sum_{cyc} \frac{x+1}{\sqrt{x^{5}+x+1}} \geq \sum_{cyc} \frac{2}{x^{2}-x+2} \geq \frac{18}{\sum_{cyc}\left(x^{2}-x+2\right)}=\frac{18}{6}=3$$
which is the desired result.
For the equality both conditions: $x^{2}-x+2=y^{2}-y+2=z^{2}-z+2$ (equality in CBS) and $x^{3}-x^{2}+1=x^{2}+x+1$ (equality in AM-GM) have to be satisfied.
By using the given condition it follows that $x^{2}-x+2+y^{2}-y+2+z^{2}-z+2=6$, hence $3\left(x^{2}-x+2\right)=6$, implying $x=0$ or $x=1$. Of these, only $x=0$ satisfies the second condition. We conclude that equality can only hold for $x=y=z=0$.
It is an immediate check that indeed for these values equality holds.

Let us present an heuristic argument to reach the key inequality $\frac{x+1}{\sqrt{x^{5}+x+1}} \geq \frac{2}{x^{2}-x+2}$.
In order to exploit the condition $x^{2}+y^{2}+z^{2}=x+y+z$ when applying CBS in Engel form, we are looking for $\alpha, \beta, \gamma>0$ such that
$$\frac{x+1}{\sqrt{x^{5}+x+1}} \geq \frac{\gamma}{\alpha\left(x^{2}-x\right)+\beta}$$
After squaring and cancelling the denominators, we get
$$(x+1)^{2}\left(\alpha\left(x^{2}-x\right)+\beta\right)^{2} \geq \gamma^{2}\left(x^{5}+x+1\right)$$
for all $x \geq 0$, and, after some manipulations, we reach to $f(x) \geq 0$ for all $x \geq 0$, where $f(x)=\alpha^{2} x^{6}-\gamma^{2} x^{5}+\left(2 \alpha \beta-2 \alpha^{2}\right) x^{4}+2 \alpha \beta x^{3}+(\alpha-\beta)^{2} x^{2}+\left(2 \beta^{2}-2 \alpha \beta-\gamma^{2}\right) x+\beta^{2}-\gamma^{2}$.
As we are expecting the equality to hold for $x=0$, we naturally impose the condition that $f$ has 0 as a double root. This implies $\beta^{2}-\gamma^{2}=0$ and $2 \beta^{2}-2 \alpha \beta-\gamma^{2}=0$, that is, $\beta=\gamma$ and $\gamma=2 \alpha$.
Thus the inequality $f(x) \geq 0$ becomes
$$\alpha^{2} x^{6}-4 \alpha^{2} x^{5}+2 \alpha^{2} x^{4}+4 \alpha^{2} x^{3}+\alpha^{2} x^{2} \geq 0, \forall x \geq 0$$
that is,
$$\alpha^{2} x^{2}\left(x^{2}-2 x-1\right)^{2} \geq 0 \forall x \geq 0$$
which is obviously true.
Therefore, the inequality $\frac{x+1}{\sqrt{x^{5}+x+1}} \geq \frac{2}{x^{2}-x+2}$ holds for all $x \geq 0$ and now we can continue as in the first solution.