Biến đổi giả thiết \(2\left(a^2+b^2\right)-\left(a+b\right)=2ab\)
Mà ta có: \(2ab\le\frac{\left(a+b\right)^2}{2}\)nên \(2\left(a^2+b^2\right)-\left(a+b\right)\le\frac{\left(a+b\right)^2}{2}\)(*)
Theo BĐT Cauchy-Schwarz: \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)nên từ (*) suy ra \(\left(a+b\right)^2-\left(a+b\right)\le\frac{\left(a+b\right)^2}{2}\)
Đặt \(s=a+b>0\)thì \(s^2-s\le\frac{s^2}{2}\Leftrightarrow\frac{s^2}{2}-s\le0\Leftrightarrow s^2-2s\le0\Leftrightarrow s\left(s-2\right)\le0\)
Mà \(s>0\)nên \(s-2\le0\Rightarrow s\le2\)hay \(a+b\le2\)
\(F=\frac{a^3}{b}+\frac{b^3}{a}+2020\left(\frac{1}{a}+\frac{1}{b}\right)\ge\frac{a^4}{ab}+\frac{b^4}{ab}+2020.\frac{4}{a+b}\)\(\ge\frac{\left(a^2+b^2\right)^2}{2ab}+\frac{8080}{a+b}\ge\left(\frac{\left(a+b\right)^2}{2}+\frac{4}{a+b}+\frac{4}{a+b}\right)+\frac{8072}{a+b}\)
\(\ge3\sqrt[3]{\frac{\left(a+b\right)^2}{2}.\frac{4}{a+b}.\frac{4}{a+b}}+\frac{8072}{2}=4042\)
Đẳng thức xảy ra khi a = b = 1
