Problem
ALG-B2-M06-P008 Maximum of r for fixed p
#8
★★★☆☆ Level 3 of 5
Let \(a,b,c\ge0\) and \(a+b+c=p\). Prove \[abc\le\frac{p^3}{27}.\]
Hint. Use AM-GM, or check that the maximum occurs at \(a=b=c\).
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Hence \(\sqrt[3]{abc}\le p/3\), so \(abc\le p^3/27\). Equality holds at \(a=b=c=p/3\).
Although this is AM-GM, the result is basic for UVW analysis.