Question

\(\frac{3x^2}{2}+y^2+z^2+yz=1\)

Tìm min, mã:

B=x+y+z

Answer 1

\(GT\Leftrightarrow3x^2+y^2+z^2+\left(y+z\right)^2=2\)

Áp dụng BĐT bunyakovsky:\(y^2+z^2\ge\frac{1}{2}\left(y+z\right)^2\)

\(2\ge\frac{3}{2}\left(y+z\right)^2+3x^2\Leftrightarrow4\ge3\left(y+z\right)^2+6x^2=3\left[\left(y+z\right)^2+2x^2\right]\)

\(\left(2+1\right)\left[\left(y+z\right)^2+2x^2\right]\ge2\left(x+y+z\right)^2\)

\(\left(x+y+z\right)^2\le2\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)