<s>[INST] Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$ 
Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$ for $a, b, c \in \mathbb{R_{>0}}$

I tried by first using AM-HM inequality on $a, b, c$ to get the following result.
$\frac{a+b+c}3 \geq \frac 3{\frac 1a+\frac1b+\frac1c}$
$\implies (a+b+c)(\frac1a+\frac1b+\frac1c) \geq 9$
$\implies (a+b+c)(ab+bc+ca) \geq 9abc$
Also I used the inequality
$(a+b)(b+c)(c+a) \geq 8abc$
But then I am not able to proceed further. Can someone please help me.
 [/INST] $$9(a+b)(b+c)(c+a) - 8(a+b+c)(ab+bc+ca)$$
$$ = (a+b)(b+c)(c+a) - 8abc$$
$$\ge0$$
because $(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca)-abc$
and $(a+b)(b+c)(c+a) \ge 8abc$ by AM-GM: $(a+b)(b+c)(c+a)=abc+\dots+bca\ge 8\sqrt[8]{a^8b^8c^8}$.
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