What does $~\sum_\text{cyclic}$ mean? I saw it in this question : Proving $~\sum_\text{cyclic}\left(\frac{1}{y^{2}+z^{2}}+\frac{1}{1-yz}\right)\geq 9$

Appended Question: Thank you @A.G for sending this link.

What is "Note that not all permutations of the variables are used; they are just cycled through" supposed to mean? How is all permutations of the variables different to what ever "cycled through" is suppose to mean? Just summing over permutation of values can be assigned to variables must be same in finite case at least.

  • 2
    $\begingroup$ Cyclic summation $\endgroup$
    – A.Γ.
    Apr 2, 2017 at 20:19
  • $\begingroup$ @A.G. : appended question $\endgroup$
    – jimjim
    Apr 2, 2017 at 21:16
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    $\begingroup$ It's pretty non-standard terminology imo, which is further motivated by A.G. having to cite AoPS rather than e.g. Wikipedia or MathWorld for its meaning. People using it should explain what it means. $\endgroup$
    – TMM
    Apr 2, 2017 at 21:28
  • $\begingroup$ @TMM: It's pretty standard terminology in IMO (and contest mathematics in general). =) $\endgroup$
    – user21820
    Apr 3, 2017 at 15:50
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    $\begingroup$ It's only (somewhat) standard in contest mathematics, not in mathematics in general. I could also tell you that you should know what PKC stands for in my field, but since the majority of people here does not work in this field and does not know what I mean, I will explain it when I use it. The same should apply when people use obscure contest math terminology. $\endgroup$
    – TMM
    Apr 4, 2017 at 1:14

2 Answers 2


There are six permutations of the variables $(x, y, z)$, namely $(x, y, z)$, $(x, z, y)$, $(y, x, z)$, $(y, z, x)$, $(z, x, y)$, and $(z, y, x)$.

There are three cyclic permutations of the variables $(x, y, z)$, namely $(x, y, z)$, $(y, z, x)$, and $(z, x, y)$.

So the sum of the expression $(x - y)^z$ over all permutations is given by

$$(x - y)^z + (x - z)^y + (y - x)^z + (y - z)^x + (z - x)^y + (z - y)^x$$

whereas the sum of the expression $(x - y)^z$ over all cyclic permutations is given by

$$(x - y)^z + (y - z)^x + (z - x)^y.$$

Note that the two sums are not the same; the first contains more summands as there are more permutations than cyclic permutations (i.e. not every permutation is a cyclic permutation, e.g. $(x, z, y)$ is a permutation of $(x, y, z)$ but not a cyclic permutation).


If you are familiar with abstract algebra, this is a special case of a more general notion.

Suppose you have a finite group $G$ acting on a space $X$ and a function $f\colon X\to {\bf R}$ (here, you can put any abelian group). Then you can consider the summation over $G$: $\sum_{g\in G} f\circ g$. This is a special case where $X={\bf R}^3$ and $G={\bf Z}_3$ acts by cyclically permuting (hence the name) the coordinates, which are $x,y$ and $z$ in the question you linked. The different case the link refers to is if you had taken instead $G=S_3$ (the full permutation group).

Frequently, this operation is paired with averaging, so we take instead $\frac{1}{\lvert G\rvert}\sum_{g\in G} f\circ g$. Note that the result of this operation is invariant under the action of $G$ and can be considered a projection of $f$ onto the space of $G$-invariant functions (because it does not change $f$ it it was $G$-invariant to begin with).

More generally, if you have a possibly infinite group $G$ with a sufficiently nice finite invariant measure $\mu$, you can do the same trick by looking at $\frac{1}{\mu(G)}\int_{G}f\circ g\,\textrm{d}\mu(g)$, again yielding an invariant function. In case of finite groups this $\mu$ is just the counting measure. This kind of averaging is a very useful idea in a variety of contexts.


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