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:

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

Corollary

Let $d$ be a non-zero real number.

Let $x$, $y$ and $z$ be real numbers with:

$x + y + z = d$

Then:

$x y + y z + z x < \dfrac {d^2} 2$

Proof

We have:

\(\ds 1\)

\(=\)

\(\ds \paren {x + y + z}^2\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {x + y} + z}^2\)

\(\ds \)

\(=\)

\(\ds \paren {x + y}^2 + 2 z \paren {x + y} + z^2\)

Square of Sum

\(\ds \)

\(=\)

\(\ds x^2 + 2 x y + y^2 + 2 x z + 2 y z + z^2\)

Square of Sum

\(\ds \)

\(=\)

\(\ds 2 \paren {x y + y z + z x} + \paren {x^2 + y^2 + z^2}\)

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:

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

$\blacksquare$

Sources

• 1980: Angela Dunn: Mathematical Bafflers (revised ed.) ... (previous) ... (next): $1$. Say it with Letters: Algebraic Amusements: An Inequality