1. Let positive real numbers $a, b$ satisfy $ab=1$. Then the maximum value of $\frac{1}{a+9b} + \frac{1}{9a+b}$ is . $\qquad$
$-1 \cdot \frac{5}{24}$.
Notice,
$$
\begin{array}{l}
\frac{1}{a+9 b}+\frac{1}{9 a+b}=\frac{10(a+b)}{9(a+b)^{2}+64} \\
=\frac{10}{9(a+b)+\frac{64}{a+b}} \\
\leqslant \frac{10}{2 \sqrt{9 \times 64}}=\frac{5}{24},
\end{array}
$$

When $a+b=\frac{8}{3}>2$, the equality holds in the above expression.