\(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)_{ }\)
\(a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)
\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)
Do đó \(P=a^2+\left(a+2\right)\left(2a\right)+2020\)
\(P=a^2+2a^2+4a+2020\)
\(P=3a^2+4a+2020\)
\(3P=9a^2+12a+6060\)
\(3P=\left(3a\right)^2+2.\left(3a\right).2+4+6060-4\)
\(3P=\left(3a+2\right)^2+6056\ge6056\Leftrightarrow3P\ge6056\Leftrightarrow P\ge\frac{6056}{3}\) Dấu "=" xảy ra khi a = b = c = \(-\frac{3}{2}\)
Vậy P đạt giá trị nhỏ nhất là 6056/3 khi a = b = c = -3/2
