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.

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.
