MathDB
- 1<=xy + yz + zt + tx <=0 if x + y + z +t = 0 and x^2+ y^2+ z^2+t^2 = 1

Source: Austrian - Polish 1996 APMC

May 3, 2020
inequalitiesalgebra

Problem Statement

Real numbers x,y,z,tx,y,z, t satisfy x+y+z+t=0x + y + z +t = 0 and x2+y2+z2+t2=1x^2+ y^2+ z^2+t^2 = 1.
Prove that 1xy+yz+zt+tx0- 1 \le xy + yz + zt + tx \le 0.