Bài 5:
\(A=x+y+z-\left(x^2+2y^2+4z^2\right)\)
\(=x+y+z-x^2-2y^2-4z^2\)
\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)-2\left(y^2-\dfrac{1}{2}y\right)-\left(4z^2-z\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}-2\left(y^2-2\cdot y\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}\right)-\left(4z^2-2\cdot2z\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}-2\left(y-\dfrac{1}{4}\right)^2+\dfrac{1}{8}-\left(2z-\dfrac{1}{4}\right)^2+\dfrac{1}{16}\)
\(=-\left(x-\dfrac{1}{2}\right)^2-2\left(y-\dfrac{1}{4}\right)^2-\left(2z-\dfrac{1}{4}\right)^2+\dfrac{7}{16}< =\dfrac{7}{16}\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y-\dfrac{1}{4}=0\\2z-\dfrac{1}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{4}\\z=\dfrac{1}{8}\end{matrix}\right.\)
