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This question is from IMO 1984 as Problem number 1 at Prague, Czechoslovakia.
Given that x, y, z are real positive integers and their sum is 1. We have to find the minimum value of xy + yz + zx - 2xyz.
Let us consider three terms, (1 - 2x), (1 - 2y), (1 - 2z).
Its product will give us terms as:
(1 - 2x)(1 - 2y)(1 - 2z)
= 1 - 2(x + y + z) + 4(xy + yz + zx) - 8xyz
= 1 - 2 + 4(xy + yz + zx) - 8xyz
= 4(xy + yz + zx) - 8xyz - 1
Now use GM ≤ AM,
[ \[\sqrt[3]{(1 - 2x)(1 - 2y)(1 - 2z)} ≤ \frac{(1 - 2x) + (1 - 2y) + (1 - 2z)}{3}\]
\[\Rightarrow\sqrt[3]{(1 - 2x)(1 - 2y)(1 - 2z)} ≤ \frac{3 - 2 (x + y + z) }{3}\]
\[\Rightarrow\sqrt[3]{(1 - 2x)(1 - 2y)(1 - 2z)} ≤ \frac{1}{3}\]
\[\Rightarrow (1 - 2x)(1 - 2y)(1 - 2z) ≤ \frac{1}{27}\]
\[\Rightarrow 4(xy + yz + zx) - 8xyz - 1 ≤ \frac{1}{27}\]
\[\Rightarrow 4(xy + yz + zx) - 8xyz ≤ \frac{28}{27}\]
\[\Rightarrow xy + yz + zx - 2xyz ≤ \frac{7}{27}\]
By this, minimum value of xy + yz + zx - 2xyz = 7/27.
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