$f(x) = n+\frac{a_1-a_2}{a_2+x} + \frac{a_2-a_3}{a_3+x} + ... + \frac{a_n-a_1}{a_1+x}$. We will prove by induction on $n$ that $f(x)$ is decreasing. Base cases $n=1, 2$ are trivial. Now assume that for some $k \ge 2$, $f(x) = n+\frac{a_1-a_2}{a_2+x} + \frac{a_2-a_3}{a_3+x} + ... + \frac{a_k-a_1}{a_1+x}$ is decreasing. We want to prove that $g(x) = n+1+\frac{a_1-a_2}{a_2+x} + \frac{a_2-a_3}{a_3+x} + ... + \frac{a_k-a_{k+1}}{a_{k+1}+x} + \frac{a_{k+1}-a_1}{a_1+x}$ is decreasing. By the inductive hypothesis, it suffices to prove that ${h(x)=\frac{a_k-a_{k+1}}{a_{k+1}+x} + \frac{a_{k+1}-a_1}{a_1+x} - \frac{a_k-a_1}{a_1+x}}$ is decreasing. Note that $h(x) = \frac{(a_{k+1}-a_k)(a_{k+1}-a_1)}{(a_1+x)(a_{k+1}+x)}$ which is decreasing, if we assume without loss of generality that $a_{k+1}$ is the greatest $a_i$.

Generalization Let $n$ be a positive integer number,$r\in \left[ 0,\infty \right)$ and let ${{a}_{1}},{{a}_{2}},\ldots ,{{a}_{n}}$ be $n$ positive real numbers. Prove that $f : [0, \infty) \rightarrow \mathbb{R}$, defined by \[f(x)=a_{1}^{r}\cdot \frac{{{a}_{1}}+x}{{{a}_{2}}+x}+a_{2}^{r}\cdot \frac{{{a}_{2}}+x}{{{a}_{3}}+x}+\cdots +a_{n-1}^{r}\cdot \frac{{{a}_{n-1}}+x}{{{a}_{n}}+x}+a_{n}^{r}\cdot \frac{{{a}_{n}}+x}{{{a}_{1}}+x},\] is a decreasing function.

Oh, but we can! The induction hypothesis is that the function on $k$ parameters is decreasing for all $k$-plets of parameters. When adding a $(k+1)$-th parameter, we can reindex so that the parameter indexed $k+1$ is indeed the largest of them, and apply the induction hypothesis to the rest of them. A quite well-known "trick" in such induction processes. EDIT. This was in answer to a previous objection by the above poster, re the induction not functioning, since we cannot assume the $(k+1)$-th parameter is the largest. After my rebuttal, the poster decided to retire his remark, so I had to explain my posting here.

ACCCGS8 wrote: $f(x) = n+\frac{a_1-a_2}{a_2+x} + \frac{a_2-a_3}{a_3+x} + ... + \frac{a_n-a_1}{a_1+x}$. We will prove by induction on $n$ that $f(x)$ is decreasing. Base cases $n=1, 2$ are trivial. Now assume that for some $k \ge 2$, $f(x) = n+\frac{a_1-a_2}{a_2+x} + \frac{a_2-a_3}{a_3+x} + ... + \frac{a_k-a_1}{a_1+x}$ is decreasing. We want to prove that $g(x) = n+1+\frac{a_1-a_2}{a_2+x} + \frac{a_2-a_3}{a_3+x} + ... + \frac{a_k-a_{k+1}}{a_{k+1}+x} + \frac{a_{k+1}-a_1}{a_1+x}$ is decreasing. By the inductive hypothesis, it suffices to prove that ${h(x)=\frac{a_k-a_{k+1}}{a_{k+1}+x} + \frac{a_{k+1}-a_1}{a_1+x} - \frac{a_k-a_1}{a_1+x}}$ is decreasing. Note that $h(x) = \frac{(a_{k+1}-a_k)(a_{k+1}-a_1)}{(a_1+x)(a_{k+1}+x)}$ which is decreasing, if we assume without loss of generality that $a_{k+1}$ is the greatest $a_i$. $h(x) = \frac{(a_{k+1}-a_k)(a_{k+1}-a_1)}{(a_1+x)(a_{k+1}+x)}$ isn't necessarily monotone decreasing.

ACCCGS8 wrote: Note that $h(x) = \frac{(a_{k+1}-a_k)(a_{k+1}-a_1)}{(a_1+x)(a_{k+1}+x)}$, which is decreasing if we assume without loss of generality that $a_{k+1}$ is the greatest $a_i$. So $h(x) = \frac{A}{(B+x)(C+x)}$, with $A \geq 0$ and $B,C > 0$, is non-increasing on $[0,\infty)$; technically we may not call it decreasing if $A=0$, since then it's constant, but that's ok.