Áp dụng Cô-si cho 3 số dương, ta có
\(\sqrt[3]{xyz}\le\frac{x+y+z}{3}=\frac{1}{3}\)
\(\Leftrightarrow xyz\le\frac{1}{27}\) (1)
\(\sqrt[3]{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\le\frac{2\left(x+y+z\right)}{3}=\frac{2}{3}\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\le\frac{8}{27}\)(2)
Từ (1),(2) =>k=\(\frac{8}{729}\)\(\Rightarrow9^3.k=8\)
Dấu "=" xảy ra khi x=y=x=1/3