A cyclic mixed cubic ratio bounds a normalized quadratic sum

Prove that for all positive real numbers $a,b$ and $c$ , $\frac {ab^{2}}{c^{2}} + \frac {bc^{2}}{a^{2}} + \frac {ca^{2}}{b^{2}} + a + b + c \ge \frac {6(a^{2} + b^{2} + c^{2})}{a + b + c} I can rewrite the inequality :$ \sum_{cyc} (a-b)^2(\frac{c}{a}-\frac{2}{a+b+c}) \geq 0$ But I don't know how to continue

Source: InternLM Lean-Workbook, record lean_workbook_12418 (Apache-2.0). Complete source proposition preserved; proof developed independently.