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.
 A: 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
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}- 2=\dfrac{(a^2+c^2
-ab-bc)^2}{(a+c)^2(ab+bc+ac)}\ge 0$$
A: 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
$4(a^2+ac+c^2)(ab+bc+ac)\le (a^2+ac+c^2+ab+bc+ac)^2
=[(a+c)^2+b(a+c)]^2
=(a+c)^2(a+b+c)^2$
A: We need to prove that
$$\frac{a^2+b^2+c^2}{ab+ac+bc}-1\geq1-\frac{8abc}{(a+b)(a+c)(b+c)}$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{2(ab+ac+bc)}\geq\frac{\sum\limits_{cyc}c(a-b)^2}{(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc}(a-b)^2(a^2b+a^2c+b^2a+b^2c-c^2a-c^2b)\geq0.$$
Let $a\geq b\geq c$.
Hence, $(a-c)^2\geq (b-c)^2$, $$a^2b+a^2c+b^2a+b^2c-c^2a-c^2b\geq0$$ and $$a^2b+a^2c+c^2a+c^2b-b^2a-b^2c\geq0.$$
Thus,
$$\sum_{cyc}(a-b)^2(a^2b+a^2c+b^2a+b^2c-c^2a-c^2b)\geq$$
$$\geq(b-c)^2(a^2b+a^2c+c^2a+c^2b-b^2a-b^2c+b^2a+b^2a+c^2a+c^2b-a^2b-a^2c)=$$
$$=2(b-c)^2c^2(a+b)\geq0.$$
Done!
