\(\Delta=b^2-4ac\le0\Rightarrow b^2\le4ac\Rightarrow\frac{a}{b}.\frac{c}{b}\ge\frac{1}{4}\)
Đặt \(\left(\frac{a}{b};\frac{c}{b}\right)=\left(x;y\right)\Rightarrow xy\ge\frac{1}{4}\)
\(F=4x+y\ge4\sqrt{xy}\ge4\sqrt{\frac{1}{4}}=2\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{1}{4}\\y=1\end{matrix}\right.\) hay \(b=c=4a\)