When is it OK to assume something like $a \ge b \ge c$ when proving inequalities? Is it okay when the inequality is cyclic instead of symmetric? For example, to prove the inequality $a^3 + b^3 + c^3 \ge a^2b + b^2c + c^2a$ (for positive real numbers $a, b, c$), can I say that WLOG, $a \ge b \ge c$?
 A: There are six possible orderings for $a,b,c$ : $a<b<c,a<c<b,\ldots,c<b<a$. (as you see each ordering corresponds to a permutation of $a,b,c$). The cyclic group $a->b->c$ acts on this set of orderings.
For each orbit under this action you can select a "distinguished representative" (the one you like best) and assume that the ordering is distinguished. In the cyclic situation, there are two orbits, so you get two cases instead of one.
A: A symmetric expression $f(a, b, c)$ means $f(a, b, c)= f(a', b', c')$ for any permutation $(a', b', c')$ of $(a, b, c)$.  Given any three real numbers, we may find a permutation that is non-decreasing, (or non-decreasing), hence we may WLOG assume $a \ge b \ge c$.
But $f(a, b, c)$ is cyclic means this works only for cyclic permutations - more precisely we have only $f(a, b, c) = f(b, c, a) = f(c, a, b)$.  There is no assurance one of the cyclic permutations will be non-increasing, so we cannot in general make the assumption $a \ge b \ge c$.  
In cyclic case we can always arrange for the first variable to be the largest (or the smallest), so in this case we may assume WLOG $a = \max(a, b, c)$ for e.g.  Or equivalently we may break it into two cases $a \ge b \ge c$ or $a \ge c \ge b$ and prove for both cases.  
A: No, we cannot. For example, consider the cyclic inequality $$a^2b+b^2c+c^2a \ge ab^2+bc^2+ca^2$$ Is true for $a \ge b \ge c$, but not true for $a \ge c \ge b$. Thus, we can only assume $a \ge b \ge c$ for symmetric inequalities.
However, we can also assume inequalities such as $a \ge b \ge c$ or $a \ge c \ge b$. If it is true for both these expressions, than it is true in general. 
A: Only when the inequality is symmetric with respect to $a,b,c$. In this case, the inequality is only cyclic, so you can assume things like $\max (a,b,c) = a$ or $\min (a,b,c) = c$. 
