L11 12 * Find real numbers $x, y$ such that $\left\{\begin{array}{l}4^{-x}+27^{-y}=\frac{5}{6}, \\ \log _{27} y-\log _{4} x \geqslant \frac{1}{6} \\ 27^{y}-4^{x} \leqslant 1\end{array}\right.$ holds.
Let $4^{x}=a, 27^{y}=b, a, b>0$, then $\left\{\begin{array}{l}\frac{1}{a}+\frac{1}{b}=\frac{5}{6}, \\ b-a \leqslant 1,\end{array}\right.$
From (1) we get $b=\frac{6 a}{5 a-6}$, substituting into (2) we get $\frac{6 a}{5 a-6}-a \leqslant 1$, which is $\frac{5 a^{2}-7 a-6}{5 a-6} \geqslant 0$. Since $a>0, b>0$, we have $5 a-6>0$, thus solving we get $a \geqslant 2$, i.e., $4^{x} \geqslant 2$, solving for $x$ we get $x \geqslant \frac{1}{2}$.

From (1) we get $a=\frac{6 b}{5 b-6}$, substituting into (2) we get $b-\frac{6 b}{5 b-6} \leqslant 1$. Since $a>0, b>0$, we have $5 b-6>0$, thus $5 b^{2}-17 b+6 \leqslant 0$, solving we get $\frac{2}{5} \leqslant b \leqslant 3$, so $\log _{27} \frac{2}{5} \leqslant y \leqslant \frac{1}{3}$.

Also, $\log _{27} y-\log _{4} x \geqslant \frac{1}{6}$, so $\log _{27} y+\frac{1}{3} \geqslant \log _{4} x+\frac{1}{2}$, from which we get $\log _{27} 3 y \geqslant \log _{4} 2 x$. Since $x \geqslant \frac{1}{2}, y \leqslant \frac{1}{3}$, we have $\log _{4} 2 x \geqslant 0, \log _{27} 3 y \leqslant 0$, hence when and only when $x=\frac{1}{2}$, $y=\frac{1}{3}$, it satisfies $\log _{27} y-\log _{4} x \geqslant \frac{1}{6}$.
In summary, $x=\frac{1}{2}, y=\frac{1}{3}$ is the unique solution.