Problem
ALG-B2-M11-P012 Roots of pairs
#12
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\ge3\sqrt{3}.\]
Hint. Compare \(\sqrt{x^2+xy+y^2}\) with \(\frac{\sqrt{3}}2(x+y)\).
Since \[x^2+xy+y^2-\frac34(x+y)^2=\frac14(x-y)^2\ge0,\] we have \(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}2(x+y)\). Summing over \((a,b)\), \((b,c)\), \((c,a)\), gives \(\frac{\sqrt{3}}2\cdot2(a+b+c)=3\sqrt{3}\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.