Áp dụng Bunhiacopski
\(\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow A\ge3^2=9\)
Vậy MinA=9
Ta có
\(\left(x^2+y^2+z^2\right)\left(1+1+1\right)\ge\left(x+y+z\right)^2\) ( Bất đẳng thức Bunyakovsky )
\(\Rightarrow\left(x^2+y^2+z^2\right)3\ge3^2\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\ge3\)
Dấu " = " xảy ra khi \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}=1\Rightarrow x=y=z=1\)
Vậy MAXA=3 khi x=y=z=1
ta có \(\frac{x^2+y^2+z^2}{3}>=\left(\frac{x+y+z}{3}\right)^{^{ }2}\)
→\(x^2+y^2+z^2>=\frac{3^2}{9}.3=3\)
dấu = xẩy ra khi x=y=z=1
vậy A min =3↔x=y=z=1
