Question

Cho hai số x , y thỏa mãn điều kiện : \(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)

Tìm GTNN của xy

Answer 1

\(2x^2+\frac{1}{x^2}+\frac{y^2}{4}=4\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}-2\right)+\left(x^2+\frac{y^2}{4}+xy\right)=2+xy\)\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(x+\frac{y}{2}\right)^2=2+xy\)

\(\Rightarrow2+xy\ge0\)\(\Rightarrow xy\ge-2\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{x}\right)^2=0\\\left(x+\frac{y}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\end{matrix}\right.\)