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) &= \dfrac{1}{2} - \dfrac{x^2 + y^2 + z^2}{2}\end{aligned}

Since x2 + y2 + z22\dfrac{x^2 + y^2 + z^2}{2} is positive, xy + yz + xz <12xy + yz + xz < \dfrac{1}{2}. That's it!