Áp dụng t/c dttsbn:
\(\dfrac{a+b+c-2020d}{d}=\dfrac{b+c+d-2020a}{a}=\dfrac{c+d+a-2020b}{b}=\dfrac{d+a+b-2020c}{c}=\dfrac{3\left(a+b+c+d\right)-2020\left(a+b+c+d\right)}{a+b+c+d}=-2017\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c-2020d=-2017d\\b+c+d-2020a=-2017a\\c+d+a-2020b=-2017b\\d+a+b-2020c=-2017c\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b+c=3d\\b+c+d=3a\\c+d+a=3b\\d+a+b=3c\end{matrix}\right.\Rightarrow a=b=c=d\)
\(F=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\\ F=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=4\)