Problem
ALG-B2-M06-P018 Schur with p equal to 1
#18
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\) and \(a+b+c=1\). Prove \[a^3+b^3+c^3+6abc\ge ab+bc+ca.\]
Hint. Rewrite \(\sum a^3\) as \(1-3q+3r\).
Since \(p=1\), \(\sum a^3=1-3q+3r\). The left side minus the right side equals \(1-4q+9r\). This is Schur degree \(3\) with \(p=1\): \(p^3-4pq+9r\ge0\). Hence the inequality holds.
Shows how a fixed sum simplifies Schur.