Cyclic sums -- How do you use them? Can someone give me an example of how cyclic sums are used? I don't really understand how they're used in problem-solving. For example, $$\sum_{a,b,c}a^2$$ Any help would be appreciated, and I'm not sure if that's the correct notation, so forgive me. Thanks for the help!
 A: Since Iuli explained what it means, just some comments to explain why we use this type of notation. 
Cyclic (and symmetric) sums occur very often in inequalities. The cyclic sum is simply the sum obtained by permuting the letters cyclical, i.e. moving the first letter to last. Same way the symmetric sum is the sum over all permutations.
They are actually extremely helpful in inequalities involving many letters, as long as you can handle them easily.
For example, lets say that you need to use in some problem the so called Muirhead inequality, in the particular case  $[7,3,2,1] > [5,4,3,1]$. Then all you write is
$$\sum_{sym} a^7b^3c^2d \geq \sum_{sym} a^5b^4c^3d $$
If you write them explicitly, each side has 24 terms and  the exact inequality is:
$$a^7b^3c^2d + a^7b^3d^2c+ a^7c^3b^2d + a^7c^3d^2b+a^7d^3c^2b+a^7d^3c^2b+
b^7a^3c^2d + b^7a^3d^2c+ b^7c^3a^2d + b^7c^3d^2a+b^7d^3c^2a+b^7d^3c^2a
+c^7b^3a^2d + c^7b^3d^2a+ c^7a^3b^2d + c^7a^3d^2b+c^7d^3a^2b+c^7d^3a^2b
c^7b^3a^2d + c^7b^3d^2a+ c^7a^3b^2d + c^7a^3d^2b+c^7d^3a^2b+c^7d^3a^2b \geq
a^5b^4c^3d + a^5b^4d^3c+ a^5c^4b^3d + a^5c^4d^3b+a^5d^4c^3b+a^5d^4c^3b+
b^5a^4c^3d + b^5a^4d^3c+ b^5c^4a^3d + b^5c^4d^3a+b^5d^4c^3a+b^5d^4c^3a
+c^5b^4a^3d + c^5b^4d^3a+ c^5a^4b^3d + c^5a^4d^3b+c^5d^4a^3b+c^5d^4a^3b
c^5b^4a^3d + c^5b^4d^3a+ c^5a^4b^3d + c^5a^4d^3b+c^5d^4a^3b+c^5d^4a^3b
$$
Which form would rather use, esspecially if it is part of a larger exercise?
Note than in general, with $n$ variables, the Muirhead inequality has n! terms on each side. So 5 variables means 240 terms, 6 variables means 1680 terms, wouldn't you rather use the half line form of it? 
There are actually few inequalities which use the cyclic notation. And even if cyclic have usually much less terms than symmetric, still they are much simpler to use in the short form.
A: Definition
Cyclic sums are used to denote summations over permutations. Consider the permutation $p=(a\, b\, c)$. The cyclic sum $\sum_{p}a$ is the sum ran through the entire permutation one cycle:
$$
\sum_p a=a+b+c.
$$
The first term is derived from the fact that the first term of a cyclic sum is always just the term being permutated. In this case, we are permutating $a$. The second term is derived from the operation of the permutation once: $a\mapsto b$. The third term is derived from the operation of the permutation twice: $a\mapsto b$ and $b\mapsto c$. In other words, each term is the permutation iterated $n-1$ times where $n$ is the position of the term in the permutation.
More rigorously, we have:
$$
\sum_p f(x_1,\ldots, x_n)=\sum_{0\le k\le n-1}p^k\left( f(x_1,\ldots, x_n)\right),
$$
for some permutation $p$ wherein $p^k$ denotes $p$ iterated $k$ times.
Example
With $p=(x_1\, \ldots \, x_n)$, the following inequality is true for even $n\le 12$ and odd $n\le 23$:
$$
\sum_p \frac{x_1}{x_2+x_3}\ge \frac{1}{2}n,
$$
where $x_i$ is nonnegative for all $x_i$ in $\{x_1,\ldots, x_n\}$ and the denominators are positive. This is referred to as Shapiro's Cyclic Sum Constant.
Also, the following paper may be of interest to you: Combinatorial Remarks on the Cyclic Sum Formula for Multiple Zeta Values.
Addendum
A personal thing I looked into was the following: Given the function $f(x)=ax^2+bx+c$ and the representation of $(s-t)$, $(r-t)$, and $(r-s)$ as $p_1$, $p_2$, and $p_3$, respectively, we can state the leading coefficient of $f$ as follows:
$$
a=\sum_{\substack{p\\ 1\le i\le 3}}f(r)\prod_{\substack{1\le j\le 3\\ j\ne i}}\frac{1}{p_i},
$$
where $p=(r\, s\, t)$ and $r$, $s$, and $t$ are $x$-coordinates of points on the parabola $f(x)$. I don't know if this is correct, as it was solely some calculation in my spare time.
For Mr. Lin:
The summation expands as follows:
\begin{align}
a&=\sum_{\substack{p\\ 1\le i\le 3}}f(r)\prod_{\substack{1\le j\le 3\\ j\ne i}}\frac{1}{p_i}\\
&=\sum_{\substack{(r\, s\, t)\\ 1\le i\le 3}}f(r)\prod_{\substack{1\le j\le 3\\ j\ne i}}\frac{1}{p_i}\\
&=f(r)\prod_{\substack{1\le j\le 3\\ j\ne 1}}\frac{1}{p_i}+f(s)\prod_{\substack{1\le j\le 3\\ j\ne 2}}\frac{1}{p_i}+f(t)\prod_{\substack{1\le j\le 3\\ j\ne 3}}\frac{1}{p_i}.
\end{align}
I am pretty sure there is a typo with the $\frac{1}{p_i}$. I believe it should be $\frac{1}{p_j}$. With that in mind:
\begin{align}
f(r)\prod_{\substack{1\le j\le 3\\ j\ne 1}}\frac{1}{p_i}+f(s)\prod_{\substack{1\le j\le 3\\ j\ne 2}}\frac{1}{p_i}+f(t)\prod_{\substack{1\le j\le 3\\ j\ne 3}}\frac{1}{p_i}&=\frac{f(r)}{p_2p_3}+\frac{f(s)}{p_1p_3}+\frac{f(t)}{p_1p_2}.
\end{align}
This notation is surely not that fun to enjoy: One must perform the permutation and the expansion upon the index set $1\le i\le 3$ simultaneously.
A: This kind of sum can be used in inequalities. 
An example of a cyclic sum is: 
$$\sum_{a,b,c}{\frac{a^3}{3}}=\frac{a^3}{3}+\frac{b^3}{3}+\frac{c^3}{3}\geq \sqrt[3]{\frac{(abc)^3}{3^3}}=\frac{abc}{3}.$$
This is a simple example but you can find a lot of them in inequality exercises.
