Question

[Hoàng Bảo Trân]

Cho \(3x^2+2y^2+2z^2+2yz=3..\)      Tìm min,max A = x+y+z

Answer 1

[Tran Le Khanh Linh]

Ta có: \(\left(x^2+y^2+2xy+2yz+2xz\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=3\)

\(\Rightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=3\)

\(\Rightarrow\left(x+y+z\right)^2\le3\)

Dấu "=" xảy ra <=> x=y=z

Do đó \(-\sqrt{3}\le x+y+z\le\sqrt{3}\)

\(\Rightarrow-\sqrt{3}\le A\le\sqrt{3}\)

=> \(\hept{\begin{cases}Min_A=-\sqrt{3}\Leftrightarrow x=y=z=\frac{-\sqrt{3}}{3}\\Max_A=\sqrt{3}\Leftrightarrow x=y=z=\frac{\sqrt{3}}{3}\end{cases}}\)