Showing that $(a^2-a+1)(b^2-b+1)(c^2-c+1) \leq 7$ How can I show that $(a^2-a+1)(b^2-b+1)(c^2-c+1) \leq 7$ given that $a+b+c = 3?$
Attempt: Setting $x=2a-1, y=2b-1, z=2c-1,$ we obtain that 
$[(2a-1)^2+3][(2b-1)^2+3][(2c-1)^2+3] \leq 448,$
$s=x+y+z = 3, q=xy+yz+zx,$ and $p=xyz.$ 
So $(x^2+y^2+z^2) = 9-2q,$ and $x^2y^2+y^2z^2+z^2x^2 = q^2-6p.$
It suffices to prove that $9(9-2q)+3(q^2-6p)+p^2+27 \leq 448,$ or equivalently, that $p^2-18p+3q^2-18q-340 \leq 0.$
Further note that $p \leq 1$ (by AM-GM).
How can I conclude?

The LHS is convex in each of the variables $a,b,c$ and $a, b,c \in [0,3]$ with $a+b+c=3$. So the LHS gets maximised when $a,b,c \in \{0,3\}$. 
It isn't hard to verify that the maximum is when any two variables are zero and the other $3$. 
