Question

Chứng minh với mọi a, b:

\(\frac{a^2+b^2+c^2}{3}\)lớn hơn hoặc bằng \(\left(\frac{a+b+c}{3}\right)^2\)

 

Answer 1

\(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)

\(\Leftrightarrow\frac{a^2+b^2+c^2}{3}\ge\frac{a^2+b^2+c^2+2ab+2bc+2ac}{9}\)

\(\Leftrightarrow\frac{3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ac}{9}\ge0\)

\(\Leftrightarrow\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{9}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{9}\ge0\) (luôn đúng  \(\forall a;b;c\) )

Vậy \(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)