3. Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$ Let $s$ $=x^{2}+y^{2}$, then $\frac{1}{s_{\text {max }}}+\frac{1}{s_{\text {min }}}$ is

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3. Solution: $\frac{8}{5}$ It is easy to know that $s=x^{2}+y^{2}>0$, let $\left\{\begin{array}{l}x=\sqrt{s} \cos \theta \\ y=\sqrt{s} \sin \theta\end{array}\right.$, 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 gives $\frac{10}{13} \leqslant s \leqslant \frac{10}{3}$, hence $S_{\max }=\frac{10}{3}, S_{\min }=$ $\frac{10}{13}$, therefore $\frac{1}{S_{\max }}+\frac{1}{S_{\min }}=\frac{3}{10}+\frac{13}{10}=\frac{8}{5}$.