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 $a_{1}+a_{2}+\cdots+a_{n}=1$ 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.
 A: 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.
A: Below is a proof for $n=3$ (which, unfortunately, does not seem to
generalize).
When $n=3$, the inequality to be shown can be rewritten as
$$
\frac{a_1}{a_2}+\frac{a_2}{a_3}+\frac{a_3}{a_1} \geq
\frac{a_1+a_3}{a_2+a_3}+\frac{a_1+a_2}{a_1+a_3}+\frac{a_2+a_3}{a_1+a_2}
\tag{1}
$$
or equivalently,
$$
\bigg(\frac{a_1}{a_2}-\frac{a_1+a_3}{a_2+a_3}\bigg)+
\bigg(\frac{a_2}{a_3}-\frac{a_1+a_2}{a_1+a_3}\bigg)+
\bigg(\frac{a_3}{a_1}-\frac{a_2+a_3}{a_1+a_2}\bigg) \geq 0
\tag{2}
$$
In other words,
$$
\bigg(\frac{a_1}{a_2}-\frac{a_1+a_3}{a_2+a_3}\bigg)+
\bigg(\frac{a_2}{a_3}-\frac{a_1+a_2}{a_1+a_3}\bigg)+
\bigg(\frac{a_3}{a_1}-\frac{a_2+a_3}{a_1+a_2}\bigg) \geq 0
\tag{3}
$$
Or
$$
\bigg(\frac{\frac{a_1}{a_2}-1}{\frac{a_2}{a_3}+1}\bigg)+
\bigg(\frac{\frac{a_2}{a_3}-1}{\frac{a_3}{a_1}+1}\bigg)+
\bigg(\frac{\frac{a_3}{a_1}-1}{\frac{a_1}{a_2}+1}\bigg) \geq 0
\tag{4}
$$
So if we put $x_k=\frac{a_k}{a_{k+1}}+1$, this is equivalent to
$$
\bigg(\frac{x_1-2}{x_2}\bigg)+
\bigg(\frac{x_2-2}{x_3}\bigg)+
\bigg(\frac{x_3-2}{x_1}\bigg) \geq 0 
\tag{5}
$$
or
$$
\frac{x_1}{x_2}+
\frac{x_2}{x_3}+
\frac{x_3}{x_1} \geq 
\frac{2}{x_1}+
\frac{2}{x_2}+
\frac{2}{x_3}
\tag{6}
$$
Now, by AM-GM we have
$$
\frac{2x_k}{x_{k+1}}+\frac{x_{k+2}}{x_k} \geq 3\bigg(\frac{x_kx_{k+2}}{x_{k+1}^2}\bigg)^{\frac{1}{3}} \tag{7}
$$
Also, Holder’s inequality implies that for any positive $w_1,w_2,w_3$,
$$
(1^3+w_1^3)(1^3+w_2^3)(1^3+w_3^3) \geq (1\times 1 \times 1+w_1w_2w_3)^3
$$
Taking $w_k=(x_k-1)^{\frac{1}{3}}=(\frac{a_k}{a_{k+1}})^{\frac{1}{3}}$, we see 
that $x_1x_2x_3 \geq 8$, and hence (7) implies that
$$
\frac{2x_k}{x_{k+1}}+\frac{x_{k+2}}{x_k} \geq \frac{6}{x_{k+1}} \tag{8}
$$
Summing on $k=1,2,3$, we deduce (6) from (8), qed.
A: This is an attempt at a full solution however there is a fatal flaw in the logic as explained in the comments by Ivan Loh and Math110.
It follows from the AM-GM inequality that the Left-Hand-Side (LHS) and the Right-Hand-Side (RHS) of the original inequality are both $\ge n$.
Consider then $\displaystyle\sum_{i=1}^n\left(\dfrac{a_i}{a_{i+1}}-1\right)-\displaystyle\sum_{i=1}^n\left(\dfrac{1-a_{i+1}}{1-a_i}-1\right) = $
$\displaystyle\sum_{i=1}^n\left(\dfrac{a_i-a_{i+1}}{a_{i+1}}-\dfrac{1-a_{i+1}-1+a_i}{1-a_i}\right) =$ 
$\displaystyle\sum_{i=1}^n\left(\dfrac{a_i-a_{i+1}}{a_{i+1}}-\dfrac{a_i-a_{i+1}}{1-a_i}\right)$
Note that $1-a_i = \displaystyle \sum_{k \ne i}a_k\gt a_{i+1}$ since all the terms of the sequence are positive.
Let $b_i = a_{i+1}-a_{i}$.  Then (thanks to robjohn for pointing this out) $\sum b_i = 0$ 
Then we have the following
$\displaystyle\sum_{i=1}^n\left(\dfrac{b_i(1-a_i)-b_ia_{i+1}}{a_{i+1}\left(1-a_i\right)}\right)$
Thanks to Math110 for pointing out my initial error.
$\displaystyle\sum_{i=1}^n\left(\dfrac{b_i-b_i(a_i + a_{i+1})}{a_{i+1}\left(1-a_i\right)}\right)$ 
$\displaystyle\sum_{i=1}^n\left(\dfrac{b_i + a_{i+1}^2 - a_i^2}{a_{i+1}\left(1-a_i\right)}\right) > \displaystyle\sum_{i=1}^n\left(\dfrac{b_i + a_{i+1}^2 - a_i^2}{\left(1-a_i\right)^2}\right) >\displaystyle\sum_{i=1}^n\left(\dfrac{b_i + a_{i+1}^2 - a_i^2}{\left(1-\min(a_i)\right)^2}\right) = 0 $ .  Thanks to Ivan Loh and Math110 for pointing out the fatal flaw in this proof attempt, we do not know if $b_i + a_{i+1}^2 - a_i^2$ is positive.  I am leaving this up here in the hopes that someone smarter than myself may find some way to extend this method into a full proof.  If it turns out this is a dead end I will happily delete this attempt at an answer.
A: let us arrange the terms in ascending order and again name it from $a_1$ to $a_n$
for $n=1$
$$\sum \frac{a_{1}}{a_{1}}=\sum \frac{1-a_{1}}{1-a_{1}}=1$$
let us assume it is true for n=k,
$$\sum \frac{a_{i}}{a_{i+1}}>\sum \frac{1-a_{i+1}}{1-a_{i}}$$
now we have to prove that it is true for $n=k+1$,i.e.,
$$\sum \frac{a_{i}}{a_{i+1}}+ \frac{a_{k+1}}{a_{1}}>\sum \frac{1-a_{i+1}}{1-a_{i}}+\frac{1-a_{1}}{1-a_{k+1}}$$
case 1:
 $a_{k+1}<a_1$
let,$a_{k+1}=q$
and $a_{1}=p$(as $a_i>0$)
then,
$$p+q<1$$
SO,
$$1-p>q$$
there fore ,
$$\frac {a_{k+1}}{a_{1}}=\frac {q}{p}$$
$$\frac {1-a_{1}}{1-a_{k+1}}=\frac {1-p}{1-q}= \frac {q+t}{p+t}$$
but we know that,
$$\frac {q+t}{p+t}<\frac{q}{p}(e.g.,3/2<2/1 and so on)$$
hence inequality holds and proved by induction,
