The positive integers $a_1,a_2, \dots, a_n$ are aligned clockwise in a circular line with $n \geq 5$. Let $a_0=a_n$ and $a_{n+1}=a_1$. For each $i \in \{1,2,\dots,n \}$ the quotient \[ q_i=\frac{a_{i-1}+a_{i+1}}{a_i} \]is an integer. Prove \[ 2n \leq q_1+q_2+\dots+q_n < 3n. \]
Problem
Source: German TSTST 2016 - #2
Tags: inequalities, Integer, number theory, number theory unsolved, n-variable inequality
rkm0959
09.07.2016 16:17
This was a Taiwan TST problem!
v_Enhance
09.07.2016 20:26
Yep. Taiwan 2014 TST. I think this was even a Taiwan MO problem in the 1990's. Really beautiful.
TelMarin
09.07.2016 20:32
Yep. And before it became a Taiwan problem, it used to be a Russian MO problem!
Diamondhead
18.07.2016 17:50
Just for a collection: Taiwan 1997
Al-Handasa
29.12.2016 23:24
Btw, the statement here is wrong : the denominators aren't a_1 but a_i.
AleksaS
18.08.2020 21:22
Woah, beautiful problem, always enjoy doing sequence problems. LHS: $$\sum_{i=1}^nq_i = \sum_{i=1}^n\frac{a_{i-1} + a_{i+1}}{a_i} = $$$$= \sum_{i=1}^n\frac{a_{i-1}}{a_i} + \sum_{i=1}^n\frac{a_i}{a_{i-1}} \geq 2n$$In last step we used $AM-GM$ inequality! RHS: We will do this by induction. Firstly we know that $$a_{i-1} + a_{i+1} = q_ia_i$$Now by summing this over all $i$ we obtain that $$\sum_{i=1}^na_{i-1} + \sum_{i=1}^na_{i+1} = \sum_{i=1}^nq_ia_i$$From this it follows that $$\sum_{i=1}^n(q_i - 2)a_i = 0$$For $q_i \geq 2$ we have that equality holds only for $q_1 = q_2 =...= q_n = 2$ for which $\sum_{i=1}^nq_i = 2n < 3n$, otherwise there must exist $q_j = 1$. Now holds that $a_j = a_1 + a_{j-1}$. Then $$\frac{\sum _{i=1}^jq_i}{j} = \frac{q_1 + q_{j-1} + q_j + \sum _{i=1}^{j-2}q_i}{j} =$$$$= \frac{3 + \frac{a_{j-2} + a_1}{a_{j-1}} + \frac{a_{2} + a_{j-1}}{a_1}}{j}$$$$< \frac{3 + 3(j-1)}{j} = 3$$This inequality completes induction.