X + y + z equals 1 implies xy + yz + zx less than Half

Theorem
Let $x$, $y$ and $z$ be real numbers such that:


 * $x + y + z = 1$

Then:


 * $x y + y z + z x < \dfrac 1 2$

Proof
We have:

So:


 * $2 \paren {x y + y z + z x} = 1 - \paren {x^2 + y^2 + z^2}$

We have:


 * $x^2 + y^2 + z^2 \ge 0$

for all real numbers $x$, $y$, $z$ with equality only if:


 * $x = y = z = 0$

This cannot be the case since $x + y + z = 1$, so we have:


 * $x^2 + y^2 + z^2 > 0$

so:


 * $2 \paren {x y + y z + z x} < 1$

giving:


 * $x y + y z + z x < \dfrac 1 2$