Problem
ALG-B2-M10-P015 Normalized Schur
#15
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[a^3+b^3+c^3+6abc\ge ab+bc+ca.\]
Hint. Write \(\sum a^3=27-9q+3r\) for \(p=3\), or use Schur.
By Schur degree \(3\), \(p^3-4pq+9r\ge0\). For \(p=3\), this is \(27-12q+9r\ge0\). Also \(\sum a^3+6r-q=(27-9q+3r)+6r-q=27-10q+9r\). Since \(q\le3\), \(27-10q+9r\ge27-12q+9r\ge0\). The result follows.
Teaching goal: choose a combination of methods and always check the equality case.