If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that $(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$ If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that
$(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$

Homogenize the inequality by writing (replace "1"s):
$$
(2a+b)(2b+c)(2c+a)+(a+b+c+a+2b)(a+b+c+b+2c)(a+b+c+c+2a) \leq 9(a+b+c)^3
$$
Evaluating the terms gives the following equivalent inequality to be shown
$$
a^3 + b^3 + c^3 - 3 abc  \ge 0
$$
Now since the variables are not restricted to nonnegative ones, this must be proved "by hand" (not by AM-GM). Observe the identity, which can be shown by multiplying everything out:
$$
a^3 + b^3 + c^3 - 3 abc  = \frac12 (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)
$$
Since $a+b+c=1$, this gives the required
$$
a^3 + b^3 + c^3 - 3 abc  = \frac12 ((a-b)^2 + (b-c)^2 + (c-a)^2) \ge 0
$$
and equality only holds for $a=b=c=\frac13$.
$\qquad \Box$
