Example 4 Given real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$, let $S=x^{2}+y^{2}$. Then $\frac{1}{S_{\text {max }}}+\frac{1}{S_{\text {min }}}=$ $\qquad$ ( $\max$ denotes the maximum value, $\min$ denotes the minimum value).
Let $x=a+b, y=a-b$. Substituting into the given equation and simplifying, we get $3 a^{2}+13 b^{2}=5$.
Therefore, $b^{2}=\frac{5}{13}-\frac{3}{13} a^{2}\left(0 \leqslant a^{2} \leqslant \frac{5}{3}\right)$.
Thus, $S=x^{2}+y^{2}=(a+b)^{2}+(a-b)^{2}$
$$
=2 a^{2}+2 b^{2}=\frac{20}{13} a^{2}+\frac{10}{13} \text {. }
$$

Therefore, when $a^{2}=0$, $S_{\text {min }}=\frac{10}{13}$;
when $a^{2}=\frac{5}{3}$, $S_{\text {max }}=\frac{20}{13} \times \frac{5}{3}+\frac{10}{13}=\frac{10}{3}$.
Hence $\frac{1}{S_{\max }}+\frac{1}{S_{\min }}=\frac{3}{10}+\frac{13}{10}=\frac{8}{5}$.