Ta có: \(3\left(a+b+c\right)^2\ge0\forall a,b,c\)
\(4\left|ab+bc+ca\right|\ge0\forall a,b,c\)
Do đó: \(3\left(a+b+c\right)^2+4\left|ab+bc+ca\right|\ge0\forall a,b,c\)
Dấu '=' xảy ra khi a=b=c=0
Ta có: \(a^{2020}-b^{2021}+c^{2022}\)
\(=0^{2020}-0^{2021}+0^{2022}\)
=0