If x + y + z = 1, show that xy + yz + xz < 0.5.
| x + y + z = 1 → | (x + y + z)2 = 1 |
| (x + y + z)(x + y + z) = 1 | |
| x2 + xy + xz + yx + y2 + yz + zx + zy + z2 = 1 | |
| x2 + y2 + z2 + 2 (xy + yz + xz) = 1. |
So 1 = x2 + y2 + z2 + 2 (xy + yz + xz)
1 – (x2 + y2 + z2) = 2 (xy + yz + xz)
½ – ((x2 + y2 + z2) ÷ 2) = xy + yz + xz.
But (x2 + y2 + z2) ÷ 2 is a positive quantity, ∴ xy + yz + xz < ½, q.e.d.