9. (16 points) Given $a, b>0, a+b=1$. Find
$$
y=\left(a+\frac{1}{2015 a}\right)\left(b+\frac{1}{2015 b}\right)
$$

the minimum value.
Given the known information, we have
\[
\begin{aligned}
y & =\left(a+\frac{1}{2015 a}\right)\left(b+\frac{1}{2015 b}\right) \\
& =a b+\frac{2015\left(a^{2}+b^{2}\right)+1}{2015^{2} a b} \\
& =a b+\frac{2016-4030 a b}{2015^{2} a b} \\
& =a b+\frac{2016}{2015^{2} a b}-\frac{2}{2015} .
\end{aligned}
\]

Let \( a b=t \in\left(0, \frac{1}{4}\right] \). Then
\[
\begin{array}{l}
y=t+\frac{2016}{2015^{2} t}-\frac{2}{2015} \\
\geqslant 2 \times \frac{\sqrt{2016}}{2015}-\frac{2}{2015}=\frac{2 \sqrt{2016}-2}{2015} .
\end{array}
\]

The equality holds if and only if \( t=\frac{\sqrt{2016}}{2015} \).
Therefore, the minimum value of \( y \) is \( \frac{2 \sqrt{2016}-2}{2015} \).