Problem
ALG-B2-M06-P019 Maximum product via two equal variables
#19
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\) and \(a+b+c=p\). Using the two-equal-variables idea, prove \(abc\le\frac{p^3}{27}\).
Hint. Check \(b=c=x\), \(a=p-2x\), and factor \(\frac{p^3}{27}-(p-2x)x^2\).
By UVW, for fixed \(p\), the extremum of \(r=abc\) occurs when two variables are equal or on the boundary. On the boundary \(r=0\). Let \(b=c=x\), \(a=p-2x\). Then \[ \frac{p^3}{27}-(p-2x)x^2=\frac{(p-3x)^2(p+6x)}{27}\ge0. \] Hence \(abc\le p^3/27\).
The same estimate as AM-GM, now through UVW logic.