Proof: Without loss of generality, let $x \geqslant y \geqslant z$, from $x+y+z=1$ we get $x \geqslant \frac{1}{3}, z = 1-x-y$.

The original inequality is equivalent to $\frac{x}{x+y(1-x-y)}+\frac{y}{y+(1-x-y) x} +\frac{1-x-y}{1-x-y+x y} \leqslant \frac{9}{4}$

Rearranging gives $\frac{x}{(x+y)(1-y)}+\frac{y}{(x+y)(1-x)}+\frac{1-x-y}{(1-x)(1-y)} \leqslant \frac{9}{4}$

Multiplying both sides of the inequality by the positive number $(1-x)(1-y)(x+y)$, it is sufficient to prove $x(1-x)+y(1-y)+(1-x-y)(x+y) \leqslant \frac{9}{4}(x+y)(x-1)(1-y)$,

Expanding, rearranging, and moving terms and arranging in descending powers of $y$, it is sufficient to prove $(9 x-1) y^{2}+(9 x-1)(x-1) y+x(1-x) \geqslant 0$,

Completing the square for $y$ on the left side of the inequality, it is sufficient to prove $(9 x-1)\left(y-\frac{1-x}{2}\right)^{2}+\frac{(3 x-1)^{2}(1-x)}{4} \geqslant 0$

Since $\frac{1}{3} \leqslant x < 1, (9 x-1) > 0, \frac{(3 x-1)^{2}(1-x)}{4} \geqslant 0$,
thus inequality (1) holds (with equality if and only if $\left\{\begin{array}{l}x=\frac{1}{3}, \\ y=\frac{1-x}{2},\end{array}\right.$ i.e., $\left\{\begin{array}{l}x=\frac{1}{3}, \\ y=\frac{1}{3}\end{array}\right.$),

Therefore, $\frac{x}{x+y z}+\frac{y}{y+z x}+\frac{z}{z+x y} \leqslant \frac{9}{4}$, with equality if and only if $x=y=z=\frac{1}{3}$.