Prove $\sum\limits_\text{cyc}\frac{a}{a+(n-1)b}\geq 1$

For $a_i> 0$, $n \in \mathbb{N}$ prove or disprove $$\frac{a_1}{a_1+(n-1) a_2} + \frac{a_2}{a_2+(n-1)a_3}+\dots+\frac{a_n}{a_n+(n-1)a_1}\geq 1.$$

Written in a cyclic notation

$$\sum_\text{cyc} \frac{a}{a+(n-1)b} \geq 1.$$

I have conjectured this one after running into couple similar ones. Perhaps it is known, but I could not find it solved anywhere. For small values of $n$ it apparently holds, for $n=1$ we have: $$\frac{a}{a} = 1 \geq 1$$

For $n=2$ it is: $$\frac{a}{a+b} + \frac{b}{b+a} = \frac{a+b}{a+b} = 1 \geq 1$$

For $n=3$ it starts to be interesting: $$\frac{a}{a+2b} + \frac{b}{b+2c} + \frac{c}{c+2a} \geq 1$$ Here the Cauchy-Schwartz inequality seems to do the trick: \begin{align} \left(\sum_\text{cyc} \frac{a} {a+2b}\right)\left(\sum_\text{cyc} a(a+2b)\right) &\geq \left(\sum_\text{cyc} a\right)^2\\ \left(\sum_\text{cyc} \frac{a}{a+2b}\right) (a+b+c)^2&\geq (a+b+c)^2 \\ \left(\sum_\text{cyc} \frac{a}{a+2b}\right) &\geq 1 \end{align}

However for higher $n$ I'm stuck. I have tried Cauchy-Schwartz inequality, as well as Holder's and Jensen's, but no luck. Also considered induction but it did not lead to anything nice.

It appears that the equality holds whenever $a_1=a_2=\cdots=a_n$. Also the inequality can be equivalently written in a form

\begin{align} \sum_\text{cyc} \frac{a}{a+(n-1)b} = \sum_\text{cyc}\frac{a+(n-1)b-(n-1)b}{a+(n-1)b} = n-(n-1)\sum_\text{cyc}\frac{b}{a+(n-1)b} &\geq 1\\ \end{align} so \begin{align} 1 &\geq \sum_\text{cyc}\frac{b}{a+(n-1)b}.\\ \end{align}

Anyone knows how to prove/disprove this for generic $n$?

• A partial answer: that is true for sure if $\frac{a_i}{a_{i+1}}\geq \frac{1}{n-1}$ holds for every $i$ (with $a_{n+1}\stackrel{\text{def}}{=}a_1$). The function $f(x)=\frac{1}{1+(n-1) e^x}$ is convex on the interval $[-\log(n-1),+\infty)$, so, under that assumption, the given inequality holds as a consequence of Jensen's inequality. – Jack D'Aurizio Dec 14 '16 at 19:56

For $n\in\{1,2\}$ our inequality is obviously true.

Let $n\geq3$ and $\frac{a_2}{a_1}=e^{x_1}$, $\frac{a_3}{a_2}=e^{x_2}$,..., $\frac{a_1}{a_n}=e^{x_n}$.

Hence, $x_1+x_2+...+x_n=0$ and we need to prove that

$\sum\limits_{i=1}^nf(x_i)\geq1$, where $f(x)=\frac{1}{1+(n-1)e^x}$.

But $f''(x)=\frac{(n-1)e^x((n-1)e^x-1)}{(1+(n-1)e^x)^3}>0$ for all $x\geq0$.

Thus, by Vasc's RCF Theorem it's enough to prove our inequality for

$e^{x_1}=e^{x_2}=...=e^{x_{n-1}}=a$ and $e^{x_n}=\frac{1}{a^{n-1}}$, which gives

$$(n-1)a\left(a^{n-1}-(n-1)a+n-2\right)\geq0,$$ which is obvious by AM-GM.

Done!

• Oh nice, didn't know about the RCF Theorem, gonna check it out. Thank you – Sil Dec 14 '16 at 21:32
• @Sil See here:artofproblemsolving.com/community/c6h32332p202112 By the way, the Vasc's proof of this theorem is very beautiful. – Michael Rozenberg Dec 14 '16 at 21:42
• @Sil It was typo. I fixed my post. Thank you! – Michael Rozenberg Dec 14 '16 at 22:15
• Awesome, managed to reproduce the result, neat theorem! Will check the proof definitely at some point :) Thanks Michael. – Sil Dec 14 '16 at 22:25