Let $\mathbb{S}_n$ be the set of all permutations of $[n]=\{1, \ldots, n\}$. For positive real numbers $d_1, \ldots, d_n$, prove $$ \sum_{\sigma \in \mathbb{S}_n} \frac{1}{d_{\sigma(1)}\left(d_{\sigma(1)}+d_{\sigma(2)}\right) \ldots\left(d_{\sigma(1)}+\cdots+d_{\sigma(n)}\right)}=\frac{1}{d_1 \ldots d_n} $$
Not sure if the below solution contains a flaw, but this is what I have tried so far:
Solution Let's define a function $g_i(x)$ for $1 \leq i \leq n$ as follows:
$$g_i(x) = \frac{x}{d_i + x}$$
Now, let's rewrite the sum in the given expression in terms of these functions:
$$S_n = \sum_{\sigma \in \mathbb{S}_n} \frac{1}{d_{\sigma(1)}\left(d_{\sigma(1)}+d_{\sigma(2)}\right) \ldots\left(d_{\sigma(1)}+\cdots+d_{\sigma(n)}\right)}$$
$$S_n = \sum_{\sigma \in \mathbb{S}_n} \frac{1}{d_{\sigma(1)}} \cdot g_{\sigma(2)}(d_{\sigma(1)}) \cdot g_{\sigma(3)}(d_{\sigma(1)}+d_{\sigma(2)}) \cdots g_{\sigma(n)}(d_{\sigma(1)}+\cdots+d_{\sigma(n-1)})$$
Due to the symmetry of the problem, we can rewrite the sum by iterating over $i$ instead of directly over the permutations:
$$S_n = \frac{1}{n!} \sum_{i_1=1}^n \sum_{i_2=1 \atop i_2 \neq i_1}^n \cdots \sum_{i_n=1 \atop i_n \neq i_{n-1}}^n \frac{1}{d_{i_1}} \cdot g_{i_2}(d_{i_1}) \cdot g_{i_3}(d_{i_1}+d_{i_2}) \cdots g_{i_n}(d_{i_1}+\cdots+d_{i_{n-1}})$$
Notice that the product inside the sums telescopes:
$$g_{i_2}(d_{i_1}) \cdot g_{i_3}(d_{i_1}+d_{i_2}) \cdots g_{i_n}(d_{i_1}+\cdots+d_{i_{n-1}}) = \frac{1}{d_{i_n}}$$
This allows us to simplify the expression further:
$$S_n = \frac{1}{n!} \sum_{i_1=1}^n \sum_{i_2=1 \atop i_2 \neq i_1}^n \cdots \sum_{i_n=1 \atop i_n \neq i_{n-1}}^n \frac{1}{d_{i_1} d_{i_n}}$$
Now, observe that each pair $(i_1, i_n)$ appears exactly $(n-2)!$ times in the sum, once for each possible permutation of the remaining $n-2$ elements. Thus, we can rewrite the sum as:
$$S_n = \frac{1}{n!} \sum_{i_1=1}^n \sum_{i_n=1 \atop i_n \neq i_1}^n (n-2)! \cdot \frac{1}{d_{i_1} d_{i_n}}$$
Finally, notice that for each fixed $i_1$, there are $(n-1)$ valid choices for $i_n$, and
Note that the factor
$$\frac{1}{d_{\sigma(1)}+d_{\sigma(2)}+\ldots+d_{\sigma(n)}}=\frac{1}{\sum_{k=1}^n d_k}$$
is common, and can be extracted in front of the sum. As for the rest, notice that the sum (with $n!$ terms) can be split into $n$ sums (of $(n-1)!$ terms), for $\sigma(n)=1$, $\sigma(n)=2$ etc. until $\sigma(n)=n$:
$$\frac{1}{\sum_{k=1}^nd_k}\times\sum_{k=1}^n\sum_{\sigma\in S_n,\\ \sigma(n)=k}\frac{1}{d_{\sigma(1)}(d_{\sigma(1)}+d_{\sigma(2)})\cdots(d_{\sigma(1)}+d_{\sigma(2)}+\ldots+d_{\sigma(n-1)})}$$
Now, we can use induction on $n$: for each $k$ in the "outer" sum, we have the "inner" sum of all permutations of $d_1, d_2,\ldots, d_n$ excluding $d_k$ (i.e., of $n-1$ numbers) which is of exactly the same form as in the original problem. Thus, each of those terms is $$\frac{1}{\underbrace{d_1d_2\cdots d_n}_{\text{excluding }d_k}}=\frac{d_k}{d_1d_2\ldots d_n}$$
as per inductive hypothesis. Let's substitute this above. What we get is:
$$\frac{1}{\sum_{k=1}^n d_k}\times\sum_{k=1}^n\frac{d_k}{d_1d_2\cdots d_n}=\frac{1}{\sum_{k=1}^n d_k}\times\frac{\sum_{k=1}^n d_k}{d_1d_2\cdots d_n}=\frac{1}{d_1d_2\cdots d_n}$$