Problem
ALG-B2-M10-P006 Two fixed estimates
#6
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[a^2+b^2+c^2+ab+bc+ca\ge6abc.\]
Hint. The left side equals \(9-(ab+bc+ca)\).
Let \(q=ab+bc+ca\), \(r=abc\). The left side equals \(9-q\). With \(a+b+c=3\), we have \(q\le3\), so \(9-q\ge6\). Also \(r\le1\), hence \(6r\le6\). Therefore \(9-q\ge6\ge6r\).
Teaching goal: choose a combination of methods and always check the equality case.