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 $a_{1}+a_{2}+\cdots+a_{n}=1$

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.

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.