Áp dụng BĐT Cosi cho 3 số dương ta có:
\(a+2b+3c\ge3\sqrt[3]{a.2b.3c}\)
\(\Rightarrow\left(a.2b.3c\right)^3\ge162abc\)
\(\Rightarrow abc\le\frac{1}{162}\)
\(a=2b=3c=\frac{1}{3}\)
Dấu " = " sảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{3}\\b=\frac{1}{6}\\c=\frac{1}{9}\end{matrix}\right.\)