# Leo Giugiuc's Cyclic Quickie in Four Variables

### Solution

Let $\displaystyle a=\frac{1}{x},$ $\displaystyle b=\frac{1}{y},$ $\displaystyle c=\frac{1}{z},$ $\displaystyle d=\frac{1}{t}.$ The required inequality becomes

$3(a+b+c+d)^2\ge 8(ab+ac+ad+bc+bd+cd)$

which is equivalent to $\displaystyle \sum_{\text{all}}(a-b)^2\ge 0,$ which is obvious.

Equality is attained for $a=b=c=d,$ i.e., $x=y=z=t.$

### Acknowledgment

The problem, with a solution, was kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page.

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