1 . 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.
(1999, Romanian Mathematical Olympiad)
Solution: Let $4^{x}=a>0, 27^{y}=b>0$. Then
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=\frac{5}{6} \\
b-a \leqslant 1
\end{array}
$$

Substituting $b \leqslant a+1$ into (1), we have
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{a+1} \leqslant \frac{5}{6} \\
5 a^{2}-7 a-6 \geqslant 0
\end{array}
$$

Solving this, we get $a \geqslant 2$, i.e., $x \geqslant \frac{1}{2}$.
Substituting $a \geqslant b-1$ into (1), when $b-1>0$, we have
$$
\begin{array}{l}
\frac{1}{b}+\frac{1}{b-1} \geqslant \frac{5}{6} \\
5 b^{2}-17 b+6 \leqslant 0
\end{array}
$$

Solving this, we get $\frac{2}{5} \leqslant b \leqslant 3$, i.e., $y \leqslant \frac{1}{3}$.
When $b<1$, we still have $y \leqslant \frac{1}{3}$.
Also, from $\log _{27} y-\log _{4} x \geqslant \frac{1}{6}$, we have
$$
\begin{array}{l}
\log _{27} y+\frac{1}{3} \geqslant \log _{4} x+\frac{1}{2}, \\
\log _{27} 3 y \geqslant \log _{4} 2 x . \\
\because x \geqslant \frac{1}{2}, y \leqslant \frac{1}{3}, \\
\therefore 2 x \geqslant 1, \log _{4} 2 x \geqslant 0 ; 3 y \leqslant 1, \log _{2} ; 3 y \leqslant 0 .
\end{array}
$$

If and only if $x=\frac{1}{2}, y=\frac{1}{3}$, it satisfies
$\log _{2} ; y^{\prime}-\log _{4} x \geqslant \frac{1}{6}$
Therefore, $x=\frac{1}{2}, y=\frac{1}{3}$.