Problem
ALG-B2-M04-P012 Dot product with zero sum
#12
★★★★☆ Level 4 of 5
Let \(a\le b\le c\), \(x\le y\le z\), \(x+y+z=0\), and \(a+b+c\ge0\). Prove that \(ax+by+cz\ge0\).
Hint. In Chebyshev, the right side contains the factor \(x+y+z\).
By Chebyshev, \[\frac{ax+by+cz}{3}\ge\frac{a+b+c}{3}\cdot\frac{x+y+z}{3}=0.\] Hence \(ax+by+cz\ge0\).
The teaching goal is to show a less obvious use of Chebyshev for determining the sign of an expression.