# work 022 – Jack Garfunkel and Murray S. Klamkin

Problem

If $xyz = (1 - x)(1 - y)(1 - z)$ where $0 \le x,y,z \le 1$ . Prove that:

$x(1 - z) + y(1 - x) + z(1 - y) \ge \frac {3}{4}$

Proof

By AM-GM,

$x(1-x) \cdot y(1-y) \cdot z(1-z) \le (\frac{1}{8})^2$

$\implies xyz \le \frac{1}{8}$

Thus ,

Set :

$x = \frac {a}{a + b}$

$y = \frac {b}{b + c}$

$z = \frac{c}{c+a}$

the inequality beomes –
$\sum_{cyc} \frac {a^2}{(a + b)(a + c)} \ge \frac {(a + b + c)^2}{a^2 + b^2 + c^2 + 3(ab + bc + ca)} \ge \frac {3}{4}$
the first inequality is C-S,
the second inequality is equivalent to –
$a^2 + b^2 + c^2 \ge ab + bc + ca$ which is true by AM-GM.
Equality occurs for $a = b = c$ or $x = y = z = \frac {1}{2}$