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

Question

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

Answer 1

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 $a=b=c=\frac13$.

$\qquad \Box$

Answer 2

Hint :

As pointed out by user Albus Dumbledore we can just consider the case where all the variable are positive now try to show that with this constraint of positivity :

$$(2a+b)(2b+c)(2c+a)\leq 1$$

And :

$$(1+a+2b)(1+b+2c)(1+c+2a)\leq 8$$