Let $x_1, x_2, \ldots ,x_n(n\ge 2)$ be real numbers greater than $1$. Suppose that $|x_i-x_{i+1}|<1$ for $i=1, 2,\ldots ,n-1$. Prove that \[\frac{x_1}{x_2}+\frac{x_2}{x_3}+\ldots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1\]
Problem
Source: Baltic Way 2010
Tags: inequalities, induction, inequalities proposed
WakeUp
19.11.2010 20:34
It is strange, the Baltic Way 2010 was held on November 6th 2010, yet this problem was posted in the Inequalities Marathon here nearly a year before. It was followed by a solution by socrates (slightly edited to strengthen $\le$ to $<$):
Use induction on $ n$. Obviously $ n\geq 2$.
For $ n=2$ our inequality writes as $ (a_1-a_2)^2\leq a_1a_2$ which is true because $ (a_1-a_2)^2\leq 1 < a_1a_2$.
Now, suppose it is true for $n$. We shall prove it is also for $n+1$.
We have
$ \displaystyle \frac {a_{1}}{a_{2}} + \frac {a_{2}}{a_{3}} + \cdots + \frac {a_{n}}{a_{n+1}}+\frac{a_{n+1}}{a_{1}}= \displaystyle \left( \frac {a_{1}}{a_{2}} + \frac {a_{2}}{a_{3}} + \cdots + \frac {a_{n}}{a_{1}}\right) +\frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}-\frac{a_n}{a_1}$
However, $ \displaystyle \frac {a_{1}}{a_{2}} + \frac {a_{2}}{a_{3}} + \cdots + \frac {a_{n}}{a_{1}}< 2n - 1$
so it is enough to show that $ \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}-\frac{a_n}{a_1}\leq 2$ or $ \frac{(a_n-a_{n+1})(a_{n+1}-a_1)}{a_1a_{n+1}}\leq 1$.
It is
$ \frac{(a_n-a_{n+1})(a_{n+1}-a_1)}{a_1a_{n+1}}\leq \frac{|(a_n-a_{n+1})||(a_{n+1}-a_1)|}{a_1a_{n+1}}$ $ \leq \left|\frac{1}{a_{n+1}}-\frac{1}{a_1}\right|< \max \left(\frac{1}{a_{n+1}},\frac{1}{a_1}\right)\leq 1$
completing the induction step.
zaya_yc
21.11.2010 15:12
WakeUp wrote: Let $x_1, x_2, \ldots x_n(n\ge 2)$ be real numbers greater than $1$. Suppose that $|x_i-x_{i+1}|<1$ for $i=1, 2,\ldots n-1$. Prove that \[\frac{x_1}{x_2}+\frac{x_2}{x_3}+\ldots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1\] $A=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\ldots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}$ $|x_i-x_{i+1}|<1$ ----> a. $x_i\geq x_{i+1}$ ---> $1\leq\frac{x_i}{x_{i+1}}<2$ <-> ($\frac{x_i-x_{i+1}}{x_{i+1}}<\frac{1}{x_{i+1}}<1$) b. $x_i\leq x_{i+1}$---> $0<\frac{x_i}{x_{i+1}}\leq1$ A---> max ---> $x_i\geq x_{i+1}$ ---> $x_1\geq x_2\geq x_3\geq...\geq x_{n}\geq1$ $\frac{x_n}{x_1}\leq1$ and $1\leq k<n-1$ --> $\frac{x_i}{x_{i+1}}<2$ $A=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\ldots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2+2+2+...+2+1=2(n-1)+1=2n-1$
Abdollahpour
21.11.2017 09:01
I think induction on $n$ solve this problem.In fact the answer of $wake up$ is correct.