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Let $a_1, a_2, \ldots, a_n $ be positive real numbers such that $\displaystyle\sum^n_{i=1} a_i=1$. Prove that

$$ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \ldots+\frac{a_n}{a^2_1+a_1}\right) \geq \frac{n}{n+1}$$

My attempt :

By Holder inequality,

$\left(\displaystyle\sum_{cyc} a_1a_2\right)\left(\displaystyle\sum_{cyc} \frac{a_1}{a^2_2+a_2}\right)\left(\displaystyle\sum_{cyc} a_1(a_2+1)\right)\geq \left(\displaystyle\sum_{cyc} a_1\right)^3$

As $\displaystyle\sum^n_{i=1} a_i=1$,

$\Leftrightarrow \left(\displaystyle\sum_{cyc}a_1a_2\right)\left(\displaystyle\sum_{cyc} \frac{a_1}{a^2_2+a_2}\right)\left(\displaystyle\sum_{cyc} a_1a_2+1\right)\geq 1$

Please suggest on how to proceed.

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  • $\begingroup$ First use CS to estimate the product and then use Jensen on $f(x)=\frac{1}{\sqrt{x+1}}$. $\endgroup$ – Frieder Jäckel Aug 20 '17 at 9:21
  • $\begingroup$ @Frieder Jäckel , I don't get used to Jensen. Can you post your answer ? I'll try to study it. $\endgroup$ – carat Aug 20 '17 at 10:44
  • $\begingroup$ Sorry there was a silly mistake in my proof. My argument doesn't work. $\endgroup$ – Frieder Jäckel Aug 20 '17 at 11:42
  • $\begingroup$ @Frieder Jäckel. No problem, thank you for your time :) $\endgroup$ – carat Aug 20 '17 at 13:12
  • $\begingroup$ I have a proof when n is odd. I try to prove it for all. $\endgroup$ – Frieder Jäckel Aug 20 '17 at 22:45
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We consider two cases:

1) Let's first consider the case when $a_1,...,a_n$ fulfill $\sum a_ia_{i+1}\leq \frac{1}{n}.$ By CS we have \begin{equation} \left(\sum a_ia_{i+1} \right)\left(\sum\frac{a_i}{a_{i+1}^2+a_{i+1}} \right)\geq \left(\sum a_i \sqrt{\frac{a_{i+1}}{a_{i+1}^2+a_{i+1}}} \right)^2=\left(\sum a_i \frac{1}{\sqrt{a_{i+1}+1}} \right)^2. \end{equation} Now observe that the function $f(x)=\frac{1}{\sqrt{x+1}}$ is clearly convex. So by Jensen \begin{equation}\sum_{i=1}^m\lambda_i f(x_i)\geq f(\sum_{i=1}^m \lambda_i x_i) \end{equation} holds for all $x_1,...,x_m>-1$ and all positive $\lambda_1,...,\lambda_m$ with $\sum_{i=1}^m\lambda_i=1.$ Applying this to $\lambda_i=a_i$ and $x_i=a_{i+1}$ we get \begin{equation} \left(\sum a_i \frac{1}{\sqrt{a_{i+1}+1}}\right)^2\geq \left(\frac{1}{\sqrt{1+\sum a_ia_{i+1}}}\right)^2=\frac{1}{1+\sum a_ia_{i+1}}. \end{equation} So to finish the proof of the inequality all we have to show is \begin{equation} n+1\geq n\left(1+\sum a_ia_{i+1} \right)=n+n\sum a_ia_{i+1}, \end{equation}which is true because $\sum a_ia_{i+1}\leq \frac{1}{n}.$

2) Now consider the case where $\sum a_i a_{i+1}\geq \frac{1}{n}.$ Because $a_1,...,a_n$ and $\frac{1}{a_1^2+a_1},...,\frac{1}{a_n^2+a_n}$ are reversly ordered, we obtain by the main theorem on ordered sequences (I hope it's called like that in english) and AM-HM \begin{equation} \sum \frac{a_i}{a_{i+1}^2+a_{i+1}}\geq \sum \frac{a_i}{a_i^2+a_i}=\sum \frac{1}{a_i+1}\geq \frac{n^2}{\sum (a_i +1)}=\frac{n^2}{n+1}. \end{equation} So in this second case \begin{equation} \left( \sum a_ia_{i+1}\right)\left( \sum \frac{a_i}{a_{i+1}^2+a_{i+1}}\right)\geq \frac{1}{n}\frac{n^2}{n+1}=\frac{n}{n+1}. \end{equation}

Observe that we don't have to prove $\sum a_ia_{i+1}\geq \frac{1}{n}$ or $\leq \frac{1}{n}$ for any such $n$ numbers $a_1,...,a_n$ with $\sum a_i=1$, but that depending on how large $\sum a_ia_{i+1}$ is, we find a proof for the inequality.

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  • $\begingroup$ May I ask where you have this inequality from? $\endgroup$ – Frieder Jäckel Aug 21 '17 at 8:44
  • $\begingroup$ Beautiful proof! +1 $\endgroup$ – Michael Rozenberg Aug 21 '17 at 9:40
  • $\begingroup$ I got this inequality from my friend's worksheet. Source is not given with the problem. I've already searched in internet but couldn't find it. BTW, I'm studying Jensen, if I don't understand any point in your solution, I'll ask you. Thanks ! $\endgroup$ – carat Aug 21 '17 at 12:59
  • $\begingroup$ Regarding 2), how does this come ? $\sum \frac{a_i}{a_{i+1}^2+a_{i+1}}\geq \sum \frac{a_i}{a_i^2+a_i}$ $\endgroup$ – carat Aug 21 '17 at 15:19
  • $\begingroup$ It is the main theorem on ordered sequences and it says: If $x_1/geq x_2\geq ... \geq x_n>0$ and $0<y_1\leq y_2 \leq ... \leq y_n$ are sequences of postivie real numbers (in that case we call the sequnces "reversly ordered"), then $\sum x_i y_{\sigma (i)}\geq \sum x_i y_i$ for any reordered sequence $(y_{\sigma (i)})$, i.e. for any bijective $\sigma :\{1,...,n\}\to\{1,...,n\}.$ $\endgroup$ – Frieder Jäckel Aug 21 '17 at 21:50

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