Problem
ALG-B2-M09-P020 Estimate through deviations
#20
★★★★★ 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 \(9-(ab+bc+ca)\). Use \(ab+bc+ca\le3\) and \(abc\le1\).
Since \(a+b+c=3\), \(\sum a^2+ab+bc+ca=(a+b+c)^2-(ab+bc+ca)=9-q\). We need \(9-q\ge6r\). From \(q\le3\), the left side is at least \(6\), and by AM-GM \(r=abc\le1\), so \(6r\le6\). Therefore \(9-q\ge6\ge6r\). Equality holds at \(a=b=c=1\).
Mixed problem: two different estimates must meet at the same equality case.