Đặt \(\left(\sqrt[3]{x};\sqrt[3]{y};\sqrt[3]{z}\right)=\left(a;b;c\right)\Rightarrow a^3+b^3+c^3=1\)
\(a^3+a^3+\frac{1}{3}\ge\frac{3a}{\sqrt[3]{3}}a^2=\sqrt[3]{9}a^2\)
Tương tự: \(2b^3+\frac{1}{3}\ge\sqrt[3]{9}b^2\); \(2c^3+\frac{1}{3}\ge\sqrt[3]{9}c^2\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+1\ge\sqrt[3]{9}\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^2+b^2+c^2\le\frac{3}{\sqrt[3]{9}}=\sqrt[3]{3}\)
\(P=ab+bc+ca\le a^2+b^2+c^2\le\sqrt[3]{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt[3]{3}}\) hay \(x=y=z=\frac{1}{3}\)