$4x^2-5xy+4y^2=19$ with $Z=x^2+y^2$, find $\dfrac1{Z_{\max}}+\dfrac1{Z_{\min}}$ I have no idea how to approach this question at all. I've tried to find the maximum and minimum of the quadratic but i am too confused on what to do afterwards.

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*From the given equation $\,xy = \cfrac{4(x^2+y^2) - 19}{5} = \cfrac{4 Z - 19}{5}\,$. But for $\forall x,y \in \mathbb{R}$ the inequality holds $\,xy \le \cfrac{1}{2}(x^2+y^2)\,$ so $\,\cfrac{4 Z - 19}{5} \le \cfrac{Z}{2} \iff \boxed{Z \le \cfrac{38}{3}}\,$

*$0 \le (x+y)^2 = x^2+y^2+2xy = Z + 2 \cfrac{4 Z - 19}{5}=\cfrac{13 Z - 38}{5} \implies \boxed{Z \ge \cfrac{38}{13}}$
$Z_{min}=\cfrac{38}{13}$ is attained for $x=-y=\sqrt{\cfrac{19}{13}}$ and $Z_{max}=\cfrac{38}{3}$ is attained for $x=y=\sqrt{\cfrac{19}{3}}\,$.
