Ta có
\(\sqrt[3]{3a3b}\le\frac{3a+3b+1}{3}\)
\(\sqrt[3]{3b3c}\le\frac{3b+3c+1}{3}\)
\(\sqrt[3]{3a3c}\le\frac{3a+3c+1}{3}\)
Cộng vế theo vế ta được
\(\sqrt[3]{9}\left(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\right)\le2\left(a+b+c\right)+1\)
<=> \(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\le\sqrt[3]{3}\)