Proof: It is known that $0 \leqslant 3 b-a<2 h$. By the lemma, we have $x^{3 b-a}+y^{3-a}+z^{3 b-a} \leqslant x^{2 b}+y^{2 b}+z^{2 b}$.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} \left(x^{a}+y^{b}+z^{b}\right)\left(x^{2 b-a}+y^{b}+z^{b}\right) \geqslant\left(x^{b}+y^{b}+z^{b}\right)^{2} . \\ \text { Then } \frac{x^{b}}{x^{a}+y^{b}+z^{b}}=\frac{x^{b}\left(x^{2 b-a}+y^{b}+z^{b}\right)}{\left(x^{a}+y^{b}+z^{b}\right)\left(x^{2 b-a}+y^{b}+z^{b}\right)} \\ \leqslant \frac{x^{3 b-a}+x^{b} y^{b}+x^{b} z^{b}}{\left(x^{b}+y^{b}+z^{b}\right)^{2}} . \end{array}$$

Similarly, $\frac{y^{b}}{y^{a}+z^{b}+x^{b}} \leqslant \frac{y^{3-a}+y^{b} z^{b}+y^{b} x^{b}}{\left(x^{b}+y^{b}+z^{b}\right)^{2}}$,
$$\frac{z^{b}}{z^{a}+x^{b}+y^{b}} \leqslant \frac{z^{3 b-a}+z^{b} x^{b}+z^{b} y^{b}}{\left(x^{b}+y^{b}+z^{b}\right)^{2}}$$

Therefore, $\frac{x^{b}}{x^{a}+y^{b}+z^{b}}+\frac{y^{b}}{y^{a}+z^{b}+x^{b}}+\frac{z^{b}}{z^{a}+x^{b}+y^{b}}$ $\leqslant \frac{\left(x^{3 b-a}+y^{3-a}+z^{3 b-a}\right)+2\left(x^{b} y^{b}+y^{b} z^{b}+z^{b} x^{b}\right)}{\left(x^{b}+y^{b}+z^{b}\right)^{2}}$
$$=\frac{\left(x^{3 b-a}+y^{3-a}+z^{3 b-a}\right)+2\left(x^{b} y^{b}+y^{b} z^{b}+z^{b} x^{b}\right)}{\left(x^{2 b}+y^{2 b}+z^{3 b}\right)+2\left(x^{b} y^{b}+y^{b} z^{b}+z^{b} x^{b}\right)}$$
$\leqslant 1$