9. Given real numbers $x, y, z$ satisfy: $x \geqslant y \geqslant z, x+y+z=1, x^{2}+y^{2}+z^{2}=3$. Find the range of real number $x$.

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According to the problem, $\left\{\begin{array}{l}y+z=1-x, \\ y^{2}+z^{2}=3-x^{2},\end{array}\right.$ By the Cauchy-Schwarz inequality, we get $(y+z)^{2} \leqslant 2\left(y^{2}+z^{2}\right)$ $\Rightarrow(1-x)^{2} \leqslant 2\left(3-x^{2}\right) \Rightarrow 3 x^{2}-2 x-5 \leqslant 0 \Rightarrow-1 \leqslant x \leqslant \frac{5}{3}$.
Similarly, we get $-1 \leqslant z \leqslant \frac{5}{3}$.
(1) $z \geqslant 0 \Rightarrow\left\{\begin{array}{l}1=x+y+z \leqslant 3 x, \\ 3=x^{2}+y^{2}+z^{2} \leqslant 3 x^{2}\end{array} \Rightarrow x \geqslant 1\right.$, in this case, the equality cannot hold;
(2) $-1 \leqslant z < 1$.

If $y < 0 \Rightarrow 2 x^{2} \geqslant x^{2}+y^{2}=3-z^{2} \geqslant 2 \Rightarrow x \geqslant 1$,
If $y \geqslant 0 \Rightarrow 2 x^{2} \geqslant x^{2}+y^{2}=3-z^{2} \geqslant 2 \Rightarrow x \geqslant 1$,
The equality holds when $(x, y, z)=(1,1,-1)$. Therefore, the range of the real number $x$ is $\left[1, \frac{5}{3}\right]$.