The inequality rewrites as $\displaystyle \sum \dfrac{(a-b)^2}{2ab} \geq 2-\dfrac{\displaystyle 2\sum ab}{3}+\prod |a-b|,$ with all sums and products being cyclic. Note that $2-\dfrac{\displaystyle 2 \sum ab}{3}=\dfrac{\displaystyle 2(3-\sum ab)}{3}=\dfrac{\displaystyle 2(\sum a^2-\sum ab)}{3}=\dfrac{\displaystyle \sum (a-b)^2}{3},$ hence we are left to prove that $\displaystyle \sum (a-b)^2(\dfrac{1}{2ab}-\dfrac{1}{3}) \geq \prod |a-b|$ However, $\dfrac{1}{2ab}-\dfrac{1}{3}=\dfrac{3-2ab}{6ab}=\dfrac{(a-b)^2+c^2}{6ab} \geq \dfrac{c|a-b|}{3ab},$ hence the desired inequality boils down to $\sum \dfrac{c|a-b|^3}{3ab} \geq \prod |a-b|,$ which readily follows due to the AM-GM inequality.

Lukaluce wrote: Let $a, b,$ and $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that $$\frac{a^2 + b^2}{2ab} + \frac{b^2 + c^2}{2bc} + \frac{c^2 + a^2}{2ca} + \frac{2(ab + bc + ca)}{3} \ge 5 + |(a - b)(b - c)(c - a)|.$$

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