2. Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$, let $S=x^{2}+y^{2}$, then the value of $\frac{1}{S_{\max }}+\frac{1}{S_{\min }}$ is
$\frac{8}{5}$
2. 【Analysis and Solution】Solution 1 It is easy to know that $s=x^{2}+y^{2}>0$, set
$$
\left[\begin{array}{l}
x=\sqrt{s} \cos \theta, \\
y=\sqrt{s} \sin \theta .
\end{array}\right.
$$

Substitute into $4 x^{2}-5 x y+4 y^{2}=5$,
we get $\sin 2 \theta=\frac{8 s-10}{5 s}$, thus $\left|\frac{8 s-10}{5 s}\right| \leqslant 1$,
which leads to $\frac{10}{13} \leqslant s \leqslant \frac{10}{3}$,
$\because S_{\max }=\frac{10}{3}, S_{\min }=\frac{10}{13}$,
$\therefore \frac{1}{S_{\max }}+\frac{1}{S_{\min }}=\frac{3}{10}+\frac{13}{10}=\frac{8}{5}$.
Solution 2 Let $x=\frac{a+b}{2}, y=\frac{a-b}{2}$, substitute into the condition to get $3 a^{2}+13 b^{2}=20, S=\frac{1}{2}\left(a^{2}+b^{2}\right)$. Therefore, we have $6 S=3$
$$
\left(a^{2}+b^{2}\right) \leqslant 3 a^{2}+13 b^{2}=20 \leqslant 26, S=13\left(a^{2}+b^{2}\right), S_{\max }=\frac{10}{13}, S_{\min }=\frac{10}{3}, \frac{1}{S_{\max }}+\frac{1}{S_{\min }}=\frac{8}{5}
$$