2110.34 – Three Numbers

If x + y + z =1x + y + z = 1, show that xy + yz + xz <0.5xy + yz + xz < 0.5.

Solution

We are given that x + y + z =1.x + y + z = 1. Therefore, (x + y + z)2 =1(x + y + z)^2 = 1. This means that:

(x+y+z)(x+y+z)=1x2+xy+xz+yx+y2+yz+zx+zy+z2=1x2+y2+z2+2(xy+yz+xz)=1(xy+yz+xz)=12−x2+y2+z22 \begin{aligned} (x + y + z) (x + y + z) &=& 1 \\ x^2 + x y + x z + y x + y^2 + y z + z x + z y + z^2 &=& 1 \\ x^2 + y^2 + z^2 + 2 (x y + y z + x z) &=& 1\\ (x y + y z + x z) &=& \frac{1}{2} - \frac{x^2 + y^2 + z^2}{2} \end{aligned}

Since (x2 + y2 + z2)/2(x^2 + y^2 + z^2)/2 is positive, xy + yz + xz <½xy + yz + xz < ½. That's it!