Example 13 (1993 National High School Mathematics League Question) Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$, let $s=x^{2}+y^{2}$, then $\frac{1}{s_{\max }}=$ $\qquad$ .
Let $x^{2}+y^{2}=t$, then considering $x, y$ as a set of data, by the variance formula, we get
$$
\begin{aligned}
s^{2} & =\frac{1}{2}\left[\left(x^{2}+y^{2}\right)-2 \cdot\left(\frac{x+y}{2}\right)^{2}\right] \\
& =\frac{1}{2}\left[\left(x^{2}+y^{2}\right)-\frac{x^{2}+2 x y+y^{2}}{2}\right] \\
& =\frac{\left(x^{2}+y^{2}\right)-2 x y}{4}=\frac{t-2 x y}{4}
\end{aligned}
$$

Given $4 x^{2}-5 x y+4 y^{2}=5$,
then $5 x y=4\left(x^{2}+y^{2}\right)-5$.
Thus, $x y=\frac{4}{5}\left(x^{2}+y^{2}\right)-1=\frac{4}{5} t-1$, substituting into (1), we get
$$
s^{2}=\frac{t-\frac{8}{5} t+2}{4}=\frac{-3 t+10}{20} \geqslant 0,
$$

Therefore, $3 t-10 \leqslant 0$, which gives $t \leqslant \frac{10}{3}$,
so $s_{\max }=\frac{10}{3}$, and thus $\frac{1}{s_{\max }}=\frac{3}{10}$.