Question

Cho \(a+b+c=2\). Tìm GTNN của \(a^2+b^2+c^2\)

Answer 1

Ta có:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\\ \Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ac\\ \Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\\ \Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\\ \Leftrightarrow3\left(a^2+b^2+c^2\right)\ge2^2=4\\ \Leftrightarrow a^2+b^2+c^2\ge\dfrac{4}{3}\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\\a+b+c=2\end{matrix}\right.\Leftrightarrow a=b=c=2.\dfrac{1}{3}=\dfrac{2}{3}\)

Vậy \(Min=\dfrac{4}{3}\Leftrightarrow a=b=c=\dfrac{2}{3}\)