<s>[INST] Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$ Let $a,b,c>0$, prove that
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$
I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.
Thanks.
 [/INST] 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
$4(a^2+ac+c^2)(ab+bc+ac)\le (a^2+ac+c^2+ab+bc+ac)^2
=[(a+c)^2+b(a+c)]^2
=(a+c)^2(a+b+c)^2$
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