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}

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