Problem
ALG-B2-M04-P015 Difference identity
#15
★★★★☆ Level 4 of 5
Let \(a\le b\le c\) and \(x\le y\le z\). Prove the identity \[3(ax+by+cz)-(a+b+c)(x+y+z)=(b-a)(y-x)+(c-a)(z-x)+(c-b)(z-y).\] Deduce Chebyshev for three terms from it.
Hint. Expand the right side and collect the coefficients of \(x,y,z\).
Expanding the right side, the coefficient of \(x\) is \(-b+a-c+a=2a-b-c\), the coefficient of \(y\) is \(b-a-c+b=-a+2b-c\), and the coefficient of \(z\) is \(c-a+c-b=-a-b+2c\). This matches the left side. Since every difference on the right is nonnegative, the left side is nonnegative, which gives Chebyshev.
A technical problem that strengthens the ability to prove Chebyshev without quoting it as a black box.