Question

Cho x,y,z \(\ge\)0 thỏa mãn \(x+y+z=\dfrac{3}{2}\).Tìm GTLN của \(A=x\sqrt[3]{y+z}+y\sqrt[3]{z+x}+z\sqrt[3]{x+y}\)

Answer 1

\(x\sqrt[3]{\left(y+z\right).1.1}\le\dfrac{x\left(y+z+1+1\right)}{3}=\dfrac{xy+xz+2x}{3}\)

\(A\le\dfrac{xy+xz+2x+yz+yx+2y+xz+zy+2z}{3}=\dfrac{2\left(xy+yz+xz\right)+2\left(x+y+z\right)}{3}=\dfrac{2}{3}\left(xy+yz+xz\right)+1\le\dfrac{2}{3}.\dfrac{\left(x+y+z\right)^2}{3}+1=\dfrac{2}{3}.\dfrac{\left(\dfrac{3}{2}\right)^2}{3}+1=\dfrac{3}{2}\)

\(\Rightarrow\)\(\)\(Max=\dfrac{3}{2}\Leftrightarrow x=y=z=\dfrac{1}{2}\)