Problem
ALG-B2-M11-P014 Third-degree Schur
#14
★★★★★ Level 5 of 5
Prove for \(a,b,c\ge0\): \[a^3+b^3+c^3+3abc\ge\sum_{\mathrm{sym}}a^2b.\]
Hint. This is Schur in the form \(\sum a(a-b)(a-c)\ge0\).
By Schur, \(\sum a(a-b)(a-c)\ge0\). Expanding gives \[a^3+b^3+c^3+3abc-\sum_{\mathrm{sym}}a^2b\ge0,\] which is exactly the claim.
Teaching goal: the student should first recognise the type of estimate, then choose the tool.