\(\sqrt{2020a+\frac{\left(b-c\right)^2}{2}}\le\sqrt{2020a+\frac{\left(b+c\right)^2}{2}}=\sqrt{2020a+\frac{\left(1010-a\right)^2}{2}}\)
\(=\sqrt{\frac{1}{2}\left(a^2+2020a+1010^2\right)}=\frac{1}{\sqrt{2}}\left(a+1010\right)\)
=> \(VT\le\frac{1}{\sqrt{2}}\left(a+b+c+3.1010\right)=2020\sqrt{2}\)
Dấu "=" xảy ra khi a=1010;b=0;c=0 và các hoán vị