TITLE: Prove $\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}\ge\sum_{i=1}^{n}\frac{1-a_{i+1}}{1-a_{i}}$ if $a_{i}>0$ and $a_{1}+a_{2}+\cdots+a_{n}=1$
QUESTION [17 upvotes]: Let $a_{i}>0,i=1,2,\cdots,n$, and $a_{1}+a_{2}+\cdots+a_{n}=1$.
How can we prove that
$$\displaystyle\sum_{i=1}^{n}\dfrac{a_{i}}{a_{i+1}}\ge\displaystyle\sum_{i=1}^{n}\dfrac{1-a_{i+1}}{1-a_{i}}$$
where $a_{n+1}=a_{1}$?
I think this can be done using the AM-GM inequality.

REPLY [8 votes]: Note that
$(1-a_i-a_{i+1})\geqslant 0$ (if $n>2$, then sign "$>$").
A).
If $a_i\geqslant a_{i+1}$, then 
$\qquad
\dfrac{a_{i}}{a_{i+1}} 
\geqslant
\dfrac{(1-a_i-a_{i+1})+a_i}{(1-a_i-a_{i+1})+a_{i+1}}
= \dfrac{1-a_{i+1}}{1-a_{i}} \geqslant 1. 
$
 
 
If $a_i<a_{i+1}$, then 
$\qquad
\dfrac{a_{i}}{a_{i+1}}
\leqslant
\dfrac{(1-a_i-a_{i+1})+a_i}{(1-a_i-a_{i+1})+a_{i+1}}
= \dfrac{1-a_{i+1}}{1-a_{i}}<1. 
$
Anyway, $\dfrac{1-a_{i+1}}{1-a_i}$ is positive number, more close to $1$, than 
$\dfrac{a_i}{a_{i+1}}$.

B).
Denote
$$
A_i = \dfrac{1-a_{i+1}}{1-a_{i}}, \qquad B_i = \dfrac{a_i}{a_{i+1}};
$$
$$
\alpha_i = \ln A_i, \qquad \beta_i = \ln B_i.
$$
We can see, that
$$
A_1\cdot A_2 \cdot \cdots \cdot A_n = B_1\cdot B_2 \cdot \cdots \cdot B_n = 1;
$$
$$
\alpha_1 + \alpha_2 + \cdots + \alpha_n = \beta_1 + \beta_2 + \cdots + \beta_n = 0;
$$
where $\;\;$ $0\leqslant \alpha_i \leqslant \beta_i$ either $\beta_i \leqslant \alpha_i\leqslant 0$, $i=1,...,n$.

C).
Then (based on here) we get $\;$
$e^{\beta_1}+\cdots+e^{\beta_n} \geqslant e^{\alpha_1}+\cdots+e^{\alpha_n}$, or (other words)
$$\sum_{i=1}^n B_i \geqslant \sum_{i=1}^n A_i.$$ Proved.