Question

[nguyễn đức sơn lâm]

 

Answer 1

[Nguyễn Việt Lâm]

Với mọi a;b;c ta luôn có:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\dfrac{9}{4}\)

\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)