Example 34 (1993 National High School League Question) Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$, let $S=$ $x^{2}+y^{2}$, then $\frac{1}{S_{\text {max }}}+\frac{1}{S_{\text {min }}}=$ $\qquad$
Solution 1: Given that the condition is a homogeneous quadratic equation, the discriminant method can be used to find the maximum and minimum values of $S$. In fact, because $x^{2} y^{2}=\left(\frac{4}{5} S-1\right)^{2}, x^{2}+y^{2}=S$, it follows that $x^{2}, y^{2}$ are the two real roots of the equation $z^{2}-S z+\left(\frac{4}{5} S-1\right)^{2}=0$. Therefore, $\Delta=S^{2}-4\left(\frac{4}{5} S-1\right)^{2}=-\frac{39}{25} S^{2}+\frac{32}{5} S-4 \geqslant 0$, solving this yields $\frac{10}{13} \leqslant S \leqslant \frac{10}{3}$. When $x=y=\pm \sqrt{\frac{5}{3}}$, $S=\frac{10}{3}$, and when $x= \pm \sqrt{\frac{5}{13}}, y=\mp \sqrt{\frac{5}{13}}$, $S=\frac{10}{13}$. Therefore, $S_{\max }=\frac{10}{3}, S_{\min }=\frac{10}{13}$, and $\frac{1}{S_{\max }}+\frac{1}{S_{\min }}=\frac{8}{5}$.

Solution 2: From the given conditions, we have $x y=\frac{4}{5} S-1$, so
$$
0 \leqslant(2 x+2 y)^{2}=4 x^{2}+8 x y+4 y^{2}=5+13 x y=5+13\left(\frac{4}{5} S-1\right)=\frac{4 \times 13}{5} S-
$$
8,
$$
0 \leqslant(2 x-2 y)^{2}=4 x^{2}-8 x y+4 y^{2}=5-3 x y=5-3\left(\frac{4}{5} S-1\right)=8-\frac{4 \times 3}{5} S \text {, }
$$

Solving this yields $\frac{10}{13} \leqslant S \leqslant \frac{10}{3}$, and the rest is the same as Solution 1.