Example 3 (1993 National High School Mathematics League Question) Real numbers $x, y$ satisfy $4x^2 - 5xy + 4y^2 = 5$, let $s = x^2 + y^2$, find $s_{\text{min}}^{1}$ + $s_{\max}^{-1}$.

---

The translation is provided as requested, maintaining the original formatting and structure.
From the given condition, we know $\frac{x^{2}+y^{2}}{s}=1$, thus, $4 x^{2}-5 x y+4 y^{2}=5 \cdot \frac{x^{2}+y^{2}}{s}$.
That is, $(4 s-5) y^{2}-5 s x y+(4 s-5) x^{2}=0$.
(*)
When $x=0$, $y^{2}=\frac{5}{4}$,
at this time, $s=x^{2}+y^{2}=-\frac{5}{4}$.
When $x \neq 0$, (*) becomes $(4 s-5)\left(\frac{y}{x}\right)^{2}-5 s \cdot \frac{y}{x}+(4 s-5)=0$, viewing this equation as a quadratic equation in $\frac{y}{x}$, then we should have
$\Delta=(5 s)^{2}-4 \cdot(4 s-5)^{2} \geqslant 0$, at this time, $\frac{10}{13} \leqslant s \leqslant \frac{10}{3}$, the equality holds under the conditions
$x=-y= \pm \frac{\sqrt{65}}{13}, x=y= \pm \frac{\sqrt{15}}{3}$.
In summary, $s_{\text {min }}^{-1}+s_{\text {max }}^{-1}=\frac{8}{5}$.