$F=tan^2\alpha-2tan\alpha+1+ \frac{1}{tan^2\alpha}-\frac{2}{tan\alpha}+1=(\frac{1}{tan^2\alpha}+\frac{tan^2\alpha}{9})+(\frac{8}{9}tan^2\alpha+\frac{8}{3})-\frac{2}{tan\alpha}-2tan\alpha-\frac{2}{3}$theo BĐT$AM-GM $$F\geq \frac{2}{3}+\frac{16}{3\sqrt{3}}tan\alpha-\frac{2}{tan\alpha}-2tan\alpha-\frac{2}{3}$ta đã biết với$\forall 6^0\leq \alpha<90^{0}$ thì $tan\alpha \sim \alpha$nên ta có $F\geq\frac{2}{3}+(\frac{16}{3\sqrt3}-2)tan60-\frac{2}{tan60}-\frac{2}{3}$$ \Rightarrow F\geq\frac{16-8\sqrt3}{3} "="\alpha=60 $

$F=tan^2\alpha-2tan\alpha+1+$$\frac{1}{tan^2\alpha}-\frac{2}{tan\alpha}+1=(\frac{1}{tan^2\alpha}+\frac{tan^2\alpha}{9})+(\frac{8}{9}tan^2\alpha+\frac{8}{3})-\frac{2}{tan\alpha}-2tan\alpha-\frac{2}{3}$theo BĐT$AM-GM $$F\geq \frac{2}{3}+\frac{16}{3\sqrt{3}}tan\alpha-\frac{2}{tan\alpha}-2tan\alpha-\frac{2}{3}$ta đã biết với$\forall 6^0\leq \alpha<90^{0}$ thì $tan\alpha \sim \alpha$nên ta có $F\geq\frac{2}{3}+(\frac{16}{3\sqrt3}-2)tan60-\frac{2}{tan60}-\frac{2}{3}$$ \Rightarrow F\geq\frac{16-8\sqrt3}{3} "="\alpha=60 $