Two inequalities from AM-GM on pairs

1. By AM-GM, $a+b\ge2\sqrt{ab}$, $b+c\ge2\sqrt{bc}$, $c+a\ge2\sqrt{ca}$. All sides are positive, so we may multiply:
$$(a+b)(b+c)(c+a)\ge8\sqrt{a^2b^2c^2}=8abc.$$
Equality iff $a=b=c$.

2. Expanding, $(a+b+c)(ab+bc+ca)=\sum_{\rm sym}a^2b+3abc$ and $(a+b)(b+c)(c+a)=\sum_{\rm sym}a^2b+2abc$, where $\sum_{\rm sym}a^2b$ has the six terms $a^2b,a^2c,b^2a,b^2c,c^2a,c^2b$. Hence
$$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc.$$
Put $P=(a+b+c)(ab+bc+ca)$. The claim becomes $P-abc\ge\frac89P$, i.e. $P\ge9abc$. By AM-GM, $a+b+c\ge3\sqrt[3]{abc}$ and $ab+bc+ca\ge3\sqrt[3]{(abc)^2}$; multiplying gives $P\ge9abc$.

Remark. Part 1 is a very frequent step: it says that a product of pairwise sums is at least $8$ times the product.

Lesson. Apply AM-GM to *small groups* of variables and multiply; and rewrite symmetric expressions with $p=a+b+c$, $q=ab+bc+ca$, $r=abc$.