Problem
ALG-B1-M07-P005 Product of three numbers
#5
★☆☆☆☆ Level 1 of 5
Let \(a,b,c>0\) and \(a+b+c=6\). Prove that \(abc\le8\).
Use AM-GM for three positive numbers.
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). The left side is \(2\), so \(\sqrt[3]{abc}\le2\), hence \(abc\le8\). Equality occurs when \(a=b=c=2\).
The equality case matters: it gives the center of the problem.