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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.

math110
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4 Answers4

8

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.

Community
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Oleg567
  • 17,295
4

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.

Ewan Delanoy
  • 61,600
0

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.

Average
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  • why is $1-a_i\gt a_{i+1}$? note that $a_i-a_{i+1}$ might not always be positive (in fact, its sum is $0$). – robjohn Mar 24 '13 at 23:32
  • @robjohn thank you for the insightful comment indeed i did not intend to claim this as a full solution but rather a partial solution that i hoped could be adapted or generalized perhaps through an additional key insight to a full proof i realize now however that i really am only considering the first n terms of the sequence (assuming they are decreasing) and not the last term which is equal to the first according to the op however maybe this approach can still be salvaged – Average Mar 24 '13 at 23:45
  • This might be something useful to look at, but keep in mind, there is nothing saying that the $a_i$ are increasing – robjohn Mar 24 '13 at 23:54
  • @robjohn i never did respond to your first question of why $ 1- a_i \gt a_{i+1}$. I apologize, the reason i think this is true is since $1-a_i$ is equal to the sum of all the other terms other than $a_i$ which includes $a_{i+1}$ and also all the terms of the sequence are positive. – Average Mar 25 '13 at 11:57
  • @math110 thank you for catching that error! I will edit the answer. Please let me know if it is ok after the edit. Thank you. – Average Mar 25 '13 at 14:09
  • @math110 Does my edited proof look ok? – Average Mar 25 '13 at 14:14
  • oh,I found have a little error,$$\dfrac{b_{i}-b_{i}(a_{i+1}+a_{i})}{a_{i+1}(1-a_{i})}=\dfrac{b_{i}-a^2_{i+1}+a^2_{i}}{a_{i+1}(1-a_{i})}$$ – math110 Mar 25 '13 at 14:22
  • Thank you very much,I think you method is very nice. – math110 Mar 25 '13 at 14:24
  • @math110 thank you for catching all those mistakes! Glad I could help. – Average Mar 25 '13 at 14:27
  • Oh, I have found this: can't konwn thins:$$b_{i+1}-a^2_{i+1}+a^2_{i}>0$$?? becasuse:$$b_{i+1}-a^2_{i+1}+a^2_{i}=a_{i+1}(1-a_{i+1})+a_{i}(a_{i}-1)$$,but $a_{i}-1<0$ – math110 Mar 25 '13 at 14:28
  • i am using $\sum b_i = 0$ and $\sum (a_i^2 - a_{i+1}^2) = 0$ This follows from direct inspection of each sum as @robjohn pointed out in an earlier comment. – Average Mar 25 '13 at 14:31
  • if:$\displaystyle\sum_{i=1}^n\left(\dfrac{b_i -a_{i+1}^2 + a_i^2}{a_{i+1}\left(1-a_i\right)}\right) > \displaystyle\sum_{i=1}^n\left(\dfrac{b_i -a_{i+1}^2 + a_i^2}{\left(1-a_i\right)^2}\right)$ ture, must:$$b_{i+1}-a^2_{i+1}+a^2_{i}>0$$,but we can't konwn this is positive – math110 Mar 25 '13 at 14:33
  • Firstly, it should be $1-a_i \geq a_{i+1}$, and not $1-a_i>a_{i+1}$. Equality can hold when $n=2$. Secondly, I think you mean $b_i=a_i-a_{i+1}$. Thirdly, $b_i+a_{i+1}^2-a_i^2$ might be negative, so $1-a_i>a_{i+1}$ may not necessarily imply that $\frac{b_i+a_{i+1}^2-a_i^2}{a_{i+1}(1-a_i)}>\frac{b_i+a_{i+1}^2-a_i^2}{(1-a_i)^2}$ – Ivan Loh Mar 25 '13 at 14:35
  • right then i go one more step and get $\displaystyle\sum_{i=1}^n\left(\dfrac{b_i + a_{i+1}^2 - a_i^2}{a_{i+1}\left(1-a_i\right)}\right) > \displaystyle\sum_{i=1}^n\left(\dfrac{b_i + a_{i+1}^2 - a_i^2}{\left(1-a_i\right)^2}\right) >\displaystyle\sum_{i=1}^n\left(\dfrac{b_i + a_{i+1}^2 - a_i^2}{\left(1-\min(a_i)\right)^2}\right)$ Each of these fractions have the same denominator hence it suffices to show that the numerators add up to 0 but this follows from @robjohn comment. – Average Mar 25 '13 at 14:35
  • No. $x>y$ does not necessarily imply that $xz>yz$. Just take $z=-1$ for a counterexample. – Ivan Loh Mar 25 '13 at 14:37
  • @IvanLoh I apologize for making such a silly mistake thank you for pointing that out, I guess this method can not be salvaged. – Average Mar 25 '13 at 14:40
  • Hello: I find this $$\left(\sum_{i=1}^{n} \dfrac{b_{i}(1-a_{i})-b_{i}a_{i+1}}{a_{i+1}(1-a_{i})}\right) =\sum_{i=1}^{n}\dfrac{-b_{i}(a_{i}+a_{i+1})}{a_{i+1}(1-a_{i})}$$ has error.because: you meaning $$\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i+1}(1-a_{i})}=0$$,but $\neq 0$ – math110 Mar 25 '13 at 14:12
  • yes, I meaning same as Ivan Loh, I think use other methods. – math110 Mar 25 '13 at 15:07
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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,

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    Sorted set is partial case. Example: ${0.1, 0.5, 0.4}$. $\sum\limits_{i=1}^3 \dfrac{a_i}{a_{i+1}} = \dfrac{0.1}{0.5}+\dfrac{0.5}{0.4} + \dfrac{0.4}{0.1}$. This sum will be changed, if you'll sort this set. – Oleg567 Jun 24 '13 at 12:38