Find the smallest constant $c$ such that for any positive integers $a_1,a_2,\ldots,a_n$ for $n \geq 3$, the following inequality holds: \begin{align} \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\cdots+\frac{a_n}{a_1+a_2}\geq cn. \end{align}

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    $\begingroup$ What have you tried? What are your thoughts? What do you think the value of $c$ is? $\endgroup$ – Calvin Lin May 11 '14 at 13:30
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    $\begingroup$ Generally not wise to put "hard" or "difficult" or other value judgements in your title. It gives the reader zero idea of the relative difficulty of your problem, because we don't know you. Titles should give the reader an idea of whether they will be able to help you, and this title simply tells me it is an inequality. $\endgroup$ – Thomas Andrews May 11 '14 at 13:43
  • $\begingroup$ Are you sure you want the smallest constant $c$? Note $c = 0$ obviously works, so does any negative $c$. $\endgroup$ – Macavity May 11 '14 at 14:13

Here is a solution for the highest $c$ that satisfies the inequality.

In general, setting all $a_i$ equal generates $LHS = n/2$. So $c=1/2$ may be a good conjecture which holds in many cases, albeit not in all. More information can be found when noticing that the inequality you propose here is Shapiro's inequality, see


This is indeed a very hard one.

As the source above states:

$c = 1/2$ if n is even and less than or equal to 12, or if n is odd and less than or equal to 23.

So for those $n$, this disproves your conjecture $c = \sqrt 2 - 1$.

As the source above further states:

For greater values of $n$, a strict (highest) lower bound is $c= \frac{\gamma}{2}$ with $\gamma \approx 0.9891\dots$. So again your $c = \sqrt 2 - 1$ is too low.

Here are some results for two particular $n$ which can be proved more easily:

For $n=3$, the inequality becomes

$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$ which is Nesbitt's inequality, this is known to have a tight lower bound at $3/2$, so $c = 1/2$.

For $n=6$, this has been treated in Prove of Nesbitt's inequality in 6 variables and a tight lower bound was established at $3$, so again $c=1/2$. However (see the comment in there) you will normally not find this as a "generalization of Nesbitt's inequality" since this terminology is used for a different inequality.


I found this problem from "Problems from the Book" by Titu Andreescu and Gabriel Dospinescu and the authors did not attempt to show the solution because of its extreme difficulty. This was solved by Vladimir Drinfeld, a Ukranian Field's medalist as the authors said.

I tried to solve it for $c=\sqrt{2}-1$ using AM-GM. but dont know how for its optimum value. \begin{align*} \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\cdots+ \frac{a_n}{a_1+a_2} \geq (\sqrt{2}-1)n \end{align*} Using Am-GM, the inequality becomes \begin{align} \frac{a_1+a_2+a_3}{a_2+a_3} \cdot \frac{a_2+a_3+a_4}{a_3+a_4}\cdots \frac{a_n+a_1+a_2}{a_1+a_2}\geq (\sqrt{2})^n \end{align}


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