A cyclic cubic ratio sum bounds eight sevenths of the total

Show that if a,b,c are positive, then $\frac {(a + b)^3}{3a^2 + 3ab + b^2} + \frac {(b + c)^3}{3b^2 + 3bc + c^2} + \frac {(c + a)^3}{3c^2 + 3ca + a^2} \ge \frac {8}{7} (a + b + c)$ follows directly from: $\frac {a^3}{3a^2 + 3ab + b^2}\geq \frac {12a - 5b}{49}$

Source: InternLM Lean-Workbook, record lean_workbook_plus_49894 (Apache-2.0). Complete source proposition preserved; proof developed independently.