Problem
ALG-B2-M11-P017 Cyclic root
#17
★★★★★ Level 5 of 5
Let \(x_1,x_2,x_3,x_4>0\), and let \(x_5=x_1\). Prove \[\sum_{i=1}^4 (x_{i+1}-x_i)\sqrt{x_i^2+3x_{i+1}^2}\ge0.\]
Hint. Prove locally that \((v-u)\sqrt{u^2+3v^2}\ge v^2-u^2\).
For \(u,v>0\), compare the root with \(u+v\). If \(v\ge u\), then \(\sqrt{u^2+3v^2}\ge u+v\). If \(v\le u\), then \(\sqrt{u^2+3v^2}\le u+v\), but \(v-u\le0\), so the direction reverses after multiplication. In both cases \[(v-u)\sqrt{u^2+3v^2}\ge(v-u)(u+v)=v^2-u^2.\] Applying this to \(u=x_i\), \(v=x_{i+1}\), the sum is at least \(\sum(x_{i+1}^2-x_i^2)=0\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.