A Cyclic Identity Forces a Value

Problem

Positive reals $x, y, z$ satisfy $xy + yz + zx \ne 1$ and

\frac{(x^2-1)(y^2-1)}{xy} + \frac{(y^2-1)(z^2-1)}{yz} + \frac{(z^2-1)(x^2-1)}{zx} = 4.

Find

\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}.

Answer

Solution

Difficulty	8/10
Topics	Substitution, algebra, Symmetry

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