\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (đúng)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}^{ }\right)^2\\ \Leftrightarrow\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{a^2+b^2+c^2+2ab+2bc+2ac}{9}\\ \Leftrightarrow\dfrac{3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2ac-2bc}{9}\ge0\\ \Leftrightarrow\dfrac{2a^2+2b^2+2c^2-2ab-2ac-2bc}{9}\ge0\\ \Leftrightarrow\dfrac{2\left(a^2+b^2+c^2\right)-2\left(ab+ac+bc\right)}{9}\ge0\)
mà ta có:
\(a^2+b^2+c^2\ge ab+bc+ac\\ \Rightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ac\right)\ge0\\ \Rightarrow\dfrac{2\left(a^2+b^2+c^2\right)}{9}\ge0\)
Vậy \(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
Giải:
Ta xét hiệu \(\dfrac{a^2+b^2+c^2}{3}-\left(\dfrac{a+b+c}{3}\right)^2\)
\(=\dfrac{1}{9}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\) \(\ge0\) (Đúng)
\(\Leftrightarrow\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\) (Đpcm)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)