Problem
ALG-B2-M10-P017 Telescoping with roots
Let \(x,y,z>0\). Prove \[(y-x)\sqrt{x^2+3y^2}+(z-y)\sqrt{y^2+3z^2}+(x-z)\sqrt{z^2+3x^2}\ge0.\]
Hint 1. Prove the local estimate \((v-u)\sqrt{u^2+3v^2}\ge v^2-u^2\).
Hint 2. Consider separately \(v\ge u\) and \(v
For \(u,v>0\), prove \((v-u)\sqrt{u^2+3v^2}\ge(v-u)(u+v)=v^2-u^2\). If \(v\ge u\), then \(\sqrt{u^2+3v^2}\ge u+v\), since \(u^2+3v^2-(u+v)^2=2v(v-u)\ge0\). If \(v
A. Source analysis. Main objects: a cyclic sum with a variable difference multiplied by a square root.
B. Insufficient first move. Squaring the whole sum directly fails because the factors have different signs.
C. Hidden observation. The root must be compared with \(u+v\), and the direction depends on the sign of \(v-u\).
D. Required move. A local estimate is proved and then summed to telescope.
E. Number of ideas. Two ideas: sign splitting and telescoping squares.
F. Difficulty justification. Regional level 7: the solution is short, but the local estimate is not obvious.
G. Why it is not one-step. Without the local estimate, the problem does not reduce to a standard formula; one must find the telescoping lower bound.