By the given condition, we have
$$1 = \frac{xyz + xy + yz + xz}{x + y + z + 1}.$$
We may therefore compute
$$\begin{aligned} 1 + x^2 &= \frac{xyz + xy + yz + xz}{x + y + z + 1} + x^2 \\ &= \frac{xyz + xy + yz + xz + x^3 + x^2y + x^2z + x^2}{x + y + z + 1} \\ &= \frac{(xyz + yz) + (x^2z + xz) + (x^2y + xy) + (x^3 + x^2)}{x + y + z + 1} \\ &= \frac{(1+x)(yz + xz + xy + x^2)}{x + y + z + 1} \\ &= \frac{(1+x)(x+y)(x+z)}{x + y + z + 1}. \end{aligned}$$
By the AM-GM inequality, this implies that
$$\frac{1}{3}\sqrt{\frac{1+x^2}{1+x}} = \frac{1}{3}\sqrt{\frac{(x+y)(x+z)}{x+y+z+1}} \le \frac{x+y+x+z}{6\sqrt{x+y+z+1}}.$$
Adding the last inequality to its cyclic analogues yields
$$\frac{1}{3} \left( \sqrt{\frac{1+x^2}{1+x}} + \sqrt{\frac{1+y^2}{1+y}} + \sqrt{\frac{1+z^2}{1+z}} \right) \le \frac{2(x+y+z)}{3\sqrt{x+y+z+1}}.$$
It thus suffices to show that
$$\frac{2(x + y + z)}{3\sqrt{x + y + z + 1}} \le \left(\frac{x + y + z}{3}\right)^{5/8}.$$
Writing $s = x + y + z$, this is equivalent to
$$\left(\frac{s}{3}\right)^{3/4} \le \frac{1}{4}(s+1) = \frac{1}{4}\left(\frac{s}{3} + \frac{s}{3} + \frac{s}{3} + 1\right),$$
which holds by the AM-GM inequality, completing the proof.