Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$

WOLG:$a\ge b\ge c$ we have $$2b(a+c)^2-(a+b)(b+c)(a+c)=(a+c)(a-b)(b-c)\ge 0$$ so $$\dfrac{8abc}{(a+b)(b+c)(a+c)}\ge\dfrac{4ac}{(a+c)^2}$$ so we only prove $$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$ since $$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}- 2=\dfrac{(a^2+c^2 -ab-bc)^2}{(a+c)^2(ab+bc+ac)}\ge 0$$

I would like to use @math 110 idea with a little difference when we prove that

$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$

Since $\dfrac{a^2+b^2+c^2}{ab+bc+ac}=\dfrac{(a+b+c)^2}{ab+bc+ac}-2$, we can rewrite the inequality as

$\dfrac{(a+b+c)^2}{ab+bc+ac}\ge 4-\dfrac{4ac}{(a+c)^2}=\dfrac{4(a^2+ac+c^2)}{(a+c)^2}$

Or $4(a^2+ac+c^2)(ab+bc+ac)\le (a+c)^2(a+b+c)^2$

Using AM-GM, we have $4(a^2+ac+c^2)(ab+bc+ac)\le (a^2+ac+c^2+ab+bc+ac)^2 =[(a+c)^2+b(a+c)]^2 =(a+c)^2(a+b+c)^2$