\[\displaystyle \sum_{\text{cyc}} \dfrac{(x-y)^2}{x+y} \geq \sqrt[3]{k \displaystyle \prod_{\text{cyc}} \dfrac{(x-y)^2}{x+y}}\]

Let \(x\), \(y\), \(z\) be non-negative real numbers such that \(xy+yz+zx\neq 0\). If the maximum value of \(k\) such that the above inequality always holds true can be represented as \(a \sqrt{b}\), where \(a\) and \(b\) are positive integers, and \(b\) is square free, find \(a+b\).

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