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Ta có:$F=\frac{a^4}{b^4}+1+\frac{b^4}{a^4}+1-(\frac{a^2}{b^2}+\frac{b^2}{a^2})+\frac{a}{b}+\frac{b}{a}-2$ $\ge2\sqrt{\frac{a^4}{b^4}.1}+2\sqrt{\frac{b^4}{a^4}.1}-(\frac{a^2}{b^2}+\frac{b^2}{a^2})+\frac{a}{b}+\frac{b}{a}-2$ $=\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a}{b}+\frac{b}{a}-2$ $=(\frac{a}{b}+\frac{b}{a})^2+\frac{a}{b}+\frac{b}{a}-4$Đặt $\frac{a}{b}+\frac{b}{a}=t\Rightarrow |t|\ge2$Xét hàm: $f(t)=t^2+t-4$ trên $(-\infty-2]\cup[2;+\infty)$ ta được: $\min_{|t|\ge2} f(t)=-2\Leftrightarrow t=-2$Vậy Min$F=2\Leftrightarrow a=-b$

Ta có:$F=\frac{a^4}{b^4}+1+\frac{b^4}{a^4}+1-(\frac{a^2}{b^2}+\frac{b^2}{a^2})+\frac{a}{b}+\frac{b}{a}-2$ $\ge2\sqrt{\frac{a^4}{b^4}.1}+2\sqrt{\frac{b^4}{a^4}.1}-(\frac{a^2}{b^2}+\frac{b^2}{a^2})+\frac{a}{b}+\frac{b}{a}-2$ $=\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a}{b}+\frac{b}{a}-2$ $=(\frac{a}{b}+\frac{b}{a})^2+\frac{a}{b}+\frac{b}{a}-4$Đặt $\frac{a}{b}+\frac{b}{a}=t\Rightarrow |t|\ge2$Xét hàm: $f(t)=t^2+t-4$ trên $(-\infty-2]\cup[2;+\infty)$ ta được: $\min_{|t|\ge2} f(t)=-2\Leftrightarrow t=-2$Vậy Min$F=-2\Leftrightarrow a=-b$