A cyclic squared-difference ratio sum with a pair-product correction

If $a,b,c$ are positive real numbers such that $a+b+c=6$ then $\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}+2(ab+bc+ca)\geq 24$ Your inequaity equivalent to $\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}+\frac{12(ab+bc+ca)}{a+b+c} \geqslant 4(a+b+c),$ or $\frac{c(a^2-2ab+bc)^2+a(b^2-2bc+ca)^2+b(c^2-2ca+ab)^2}{abc(ab+bc+ca)} \geqslant 0.$ Done.

Source: InternLM Lean-Workbook, record lean_workbook_21064 (Apache-2.0). Complete source proposition preserved; proof developed independently.