# Complicated Inequality

Algebra Level 5

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

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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