Đặt \(\left(\sqrt[3]{x+y};\sqrt[3]{y+z};\sqrt[3]{z+x}\right)=\left(a;b;c\right)\Rightarrow a^3+b^3+c^3=2\)
\(S=a+b+c\)
Ta có: \(a^3+\frac{2}{3}+\frac{2}{3}\ge3\sqrt[3]{\frac{4}{9}}a=\sqrt[3]{12}a\)
\(b^3+\frac{2}{3}+\frac{2}{3}\ge\sqrt[3]{12}b\) ; \(c^3+\frac{2}{3}+\frac{2}{3}\ge\sqrt[3]{12}c\)
\(\Rightarrow\sqrt[3]{12}\left(a+b+c\right)\le a^3+b^3+c^3+4=6\)
\(\Rightarrow a+b+c\le\frac{6}{\sqrt[3]{12}}=\sqrt[3]{18}\)
\(S_{max}=\sqrt[3]{18}\) khi \(a=b=c=\sqrt[3]{\frac{2}{3}}\) hay \(x=y=z=\frac{1}{3}\)
