I have difficulties in evaluating the multi-index notation in the following context:
Let $x \in R^n$ and let $i$ be a multi-index, $i=(i_1, \dots, i_n)$. Now I want to know the bound of the sum (knowing that each $x_i \leq a$, $\frac{1}{x_i} \leq b$) $ \sum_{|i|=3} \frac{1}{i!} \frac{x^i}{x_l} \quad (\ast) $
What I thought so far:
There are three types of summands, (i) there is one $i_j=3$ (and the rest $i_k$, $k \neq j$ is 0) in $i$, then $i!$=6; (ii) there is one index $i_{j_1}=2$ and one $i_{j_2}$=1 (the others 0), then $i!=2$; and (iii) there are three components of $i$ 1, the rest 0, the faculty would be 1. (the types would be $(0, \dots, 3, \dots, 0)$, $(0, \dots, 0 , 2, 0, \dots, 0, 1, 0, \dots, 0)$ and $(0, \dots, 0 , 1, 0, \dots, 0, 1, 0, \dots, 0, 1, 0, \dots, 0)$).
Now if I look at the first case, I would have $(n-1)$ times the expression $\frac{x_j^3}{x_l}$, $l \neq j$ and once $x_l^2$, so $ (\ast) = \frac{1}{6} ( \underbrace{\sum_{j \neq l} \frac{x_j^3}{x_l}}_{ \leq (n-1) a^3 b} \quad + \underbrace{x_l^2}_{\leq a^2}) + \frac{1}{2} (\text{???}) + \frac{1}{1} (\text{????}) $
Q: How can I obtain the whole bound for $(\ast)$ ?
Is this right so far? How can I continue for the second case (ii)? Is there a way to let it do mechanically be mathematica? (Does someone know how to put multi-index notation in Mathematica)
Note: I write $x_i \leq a$, $\frac{1}{x_i} \leq b$ because in the application I look at $E(\frac{x^i}{x_l})$ and have bounds for the expectation values.