Áp dụng BĐT Bun-hia-cop-xki ta có:
\(\left(a^2+b^2+c^2\right)\left(1^2+1^2+1^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\frac{4}{3}\)
Dấu '=' xảy ra khi \(\hept{\begin{cases}a=b=c\\a+b+c=2\end{cases}\Leftrightarrow a=b=c=\frac{2}{3}}\)
Vậy \(A_{min}=\frac{4}{3}\)khi \(a=b=c=\frac{2}{3}\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
Suy ra \(A=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\)
\(=4-2\left(ab+bc+ca\right)\)
Ta có BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\).Thay vào tìm được min
