The first inequality is equivalent to
$$\frac{z^2}{x^2 + y^2} \leq z\sqrt{1-z^2}.$$
Since
$$z\sqrt{1-z^2} = \sqrt{z^2(1-z^2)} \leq \frac{z^2 + 1 - z^2}{2} = \frac{1}{2},$$
it follows that $\frac{z^2}{x^2 + y^2} \leq 2$, and therefore $x^2 + y^2 \geq 2z^2$. Writing the other two similar inequalities and adding them together yields $2(x^2 + y^2 + z^2) \geq 2(x^2 + y^2 + z^2)$. Consequently, equality must hold in all the inequalities above, hence $x = y = z = \frac{\sqrt{2}}{2}$. Clearly this triple satisfies all the requirements.

As $x, y, z \in (0,1)$, we can write $x = \sin \alpha$, $y = \sin \beta$, $z = \sin \gamma$, with $\alpha, \beta, \gamma \in (0, \pi/2)$. The first inequality becomes
$$\sin^2 \alpha + \sin^2 \beta \geq \tan \gamma \geq 2 \sin^2 \gamma,$$
because it reduces to $\sin \gamma(2 \sin \gamma \cos \gamma - 1) = \sin \gamma(\sin 2\gamma - 1) \leq 0$. Adding this inequality with the two similar ones obtained from the second and third condition, we get $2 \sum \sin^2 \alpha \geq 2 \sum \sin^2 \alpha$. Consequently, equality must hold in all the preceding inequalities, therefore $\alpha = \beta = \gamma = \pi/4$, which translates into $x = y = z = \sqrt{2}/2$.