\(\frac{\left(a+b+c\right)^2}{3}\ge ab+ac+bc\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3ab+3ac+3bc\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3ab+3ac+3bc\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
Vậy ta có đpcm