TA có: \(3\left(a^2+b^2+c^2\right)-\left(a+b+c\right)^2\)
\(=3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2ac-2bc\)
\(=2a^2+2b^2+2c^2-2ab-2ac-2bc\)
\(=\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2+c^2-2bc\right)=\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\forall a,b,c\)
=>\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\) (1)
Ta có: \(\left(a+b+c\right)^2-3\left(ab+ac+bc\right)\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-3ab-3ac-3bc\)
\(=a^2+b^2+c^2-ab-ac-bc=\frac12\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\)
\(=\frac12\cdot\left\lbrack\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2+c^2-2bc\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right\rbrack>=0\forall a,b,c\)
=>\(\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\) (2)
Từ (1),(2) suy ra \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)
