Cyclic Inequality in n (at least 4) variables Proof of a cyclic inequality.
Let $a_i$ , $i=1..n$, $n \geq 3$ be real numbers with $a_i > 0 $ and $\prod_{i=1}^n a_i= 1$. Denote $S = \sum_{i=1}^n a_i$ (unrestricted).   
Prove (or disprove) the following cyclic (i.e. $a_{n+1} = a_{1}$) inequality at least for $n=4$, preferably for general $n \geq 6$: 
$$\sum_{i=1; cyc}^n  \frac{a_i \; \{S - (a_i + a_{i+1})\}}{a_i + a_{i+1}} \geq \frac{(n-2)\, n }{2}$$
Remarks:


*

*Obviously, if all $a_i$ are equal, $a_i = a$, we have $a = 1$ due to the product condition and hence one gets equality since 
$$LHS = \sum_{i=1; cyc}^n  \frac{n -2}{2} = \frac{(n -2)n}{2}$$

*The case for $n=3$ is given by 
$$ \frac{a\; c}{a +b} + \frac{b\; a}{b +c} + \frac{c\; b}{c +a} \geq \frac{3}{2}$$
or equivalently, due to the product condition,
$$ \frac{1}{b(a +b)} + \frac{1}{c(b +c)} + \frac{1}{a(c +a)} \geq \frac{3}{2}$$ 
which is known (2008 International Zhautykov Olympiad), for some proofs see here: http://artofproblemsolving.com/community/c6h183916p1010959 


*In the meantime (29. April 2016), also the case $n=5$ was proven (take into account the comment below by Macavity Mar 25 at 6:36) by the author, see here: Inequality with five variables
 A: Here is a proof for general  $n$. 
As pointed out in the comment by  Macavity on Mar 25 at 6:36,  the claim is equivalent to:
$$\sum_{cyc}\frac{a_i}{a_i+a_{i+1}}\geq 1+\frac{n(n-2)}{2S}$$
We can remove the requirement $\prod_{i=1}^n a_i=1$ by homogenizing  this to (simultaneously defining $L$ and $R$):
$$L = \sum_{cyc}\frac{a_i}{a_i+a_{i+1}}\geq 1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} = R$$
The discussion following here is a generalization of the discussion for $n=5$ at Inequality with five variables
We write $2 L \geq 2 R$ or $L \geq 2 R- L$ and add on both sides a term $$\sum_{cyc}\frac{a_{i+1}}{a_i+a_{i+1}}$$ which leaves us to show
$$n = \sum_{cyc}\frac{a_i + a_{i+1}}{a_i+a_{i+1}}\geq 2+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{S} + \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}}$$
or, in reformulation of our claim, 
$$ \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} \leq (n -  2) (1- \frac{n \sqrt[n]{a_1a_2...a_n}}{S} )$$
We will need the following Lemma (required below), which is the above L-R-inequality for 3 variables:
$$  \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c} \leq  (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )$$
This Lemma is, from the above discussion, just a re-formulation of the claim in $L$ and $R$ above, for 3 variables, i.e.
$$  \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq  1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$$
By homogeneity, we can demand $abc=1$ and under that restriction, we need to show   $$  \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq  1+\frac{3}{2(a+b+c)}$$
This reformulates into 
 $$ \frac{a\; c}{a +b} + \frac{b\; a}{b +c} + \frac{c\; b}{c +a} \geq \frac{3}{2}$$
which is known (see remark 2 in the problem description).
Hence the Lemma holds.
For general $n$, we rewrite the LHS of our above reformulation by adding and subtracting terms $\frac{a_{k}-a_1}{a_{k}+a_1}$, for $k=3 \ldots (n-1)$. Here, the first terms are given by $a,b,c,d,e$ for better readability.
$$ \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} = \\
(\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + (\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e})
\\
+ \cdots + (\frac{a_{n-1}-a_1}{a_{n-1}+a_1}+ \frac{a_{n}-a_{n-1}}{a_{n}+a_{n-1}} + \frac{a_1-a_n}{a_1+a_n})
$$ 
There are in total $n-2$ of these terms with three summands each. 
This also holds for any of the $n$ cyclic shifts in $(a_1 \cdots a_n)$, denoted by $cyc(n)$, so we can write
$$ n \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} = \\
\sum_{cyc (n)} (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + \sum_{cyc (n)}(\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + \sum_{cyc (n)} (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e}) +
\cdots +
\sum_{cyc (n)} (\frac{a_{n-1}-a_1}{a_{n-1}+a_1}+ \frac{a_{n}-a_{n-1}}{a_{n}+a_{n-1}} + \frac{a_1-a_n}{a_1+a_n})
$$ 
Since all triples of terms are in 3 variables only, we can use our Lemma. Then  it suffices to show 
$$
\sum_{cyc (n)} (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )
 + \sum_{cyc (n)}(1- \frac{3 \sqrt[3]{a\, c \, d}}{a + c+ d} )
+ \sum_{cyc (n)}(1- \frac{3 \sqrt[3]{a\, d \, e}}{a + d+ e} )
+ \cdots
+ \sum_{cyc (n)}(1- \frac{3 \sqrt[3]{a_1\, a_{n-1} \, a_n}}{a_1 + a_{n-1} + a_n} )\\
\leq n (n-2) (1- \frac{n \sqrt[n]{a_1a_2...a_n}}{S} )
$$
which is
$$
\sum_{cyc (n)} (\frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )
 + \sum_{cyc (n)}(\frac{3 \sqrt[3]{a\, c \, d}}{a + c+ d} )
+ \sum_{cyc (n)}(\frac{3 \sqrt[3]{a\, d \, e}}{a + d+ e} )
+ \cdots
+ \sum_{cyc (n)}(\frac{3 \sqrt[3]{a_1\, a_{n-1} \, a_n}}{a_1 + a_{n-1} + a_n} )\\
\geq n^2 \, (n-2) (\frac{\sqrt[n]{a_1a_2...a_n}}{S} )
$$
Using Cauchy-Schwarz leaves us with showing
$$
\frac {(\sum_{cyc (n)} \sqrt[6]{a\, b \, c})^2}{\sum_{cyc (n)}(a + b+ c)} 
+
\frac {(\sum_{cyc (n)} \sqrt[6]{a\, c \, d})^2}{\sum_{cyc (n)}(a + c+ d)} 
+
\frac {(\sum_{cyc (n)} \sqrt[6]{a\, d \, e})^2}{\sum_{cyc (n)}(a + d+ e)}
+ \cdots +
\frac {(\sum_{cyc (n)} \sqrt[6]{a_1\, a_{n-1} \, a_n})^2}{\sum_{cyc (n)}(a_1 + a_{n-1} + a_n)} 
\geq \frac{1}{3} n^2 \,(n-2)  \frac{\sqrt[n]{a_1a_2...a_n}}{S}
$$
The denominators all equal $3S$, so this becomes
$$ {(\sum_{cyc (n)} \sqrt[6]{a\, b \, c})^2} 
+{(\sum_{cyc (n)} \sqrt[6]{a\, c \, d})^2} 
+{(\sum_{cyc (n)} \sqrt[6]{a\, d \, e})^2}
+ \cdots + {(\sum_{cyc (n)} \sqrt[6]{a_1\, a_{n-1} \, a_n})^2}\\
\geq n^2 \,(n-2)  \sqrt[n]{a_1a_2...a_n}
$$
Using AM-GM gives for the first term
$$
(\sum_{cyc (n)} \sqrt[6]{a\, b \, c})^2 \geq ( n (\prod_{cyc (n)} \sqrt[6]{a\, b \, c} )^{1/n})^2 
= n^2 (\prod_{cyc (n)} ({a\, b \, c} ) )^{1/(3 n)}  = n^2 \sqrt[n]{a_1a_2...a_n}
$$
By the same procedure, the other  term on the LHS are likewise greater or equal than $n^2 \sqrt[n]{a_1a_2...a_n}$. There are $n-2$ of these terms, hence the sum on the LHS after AM-GM equals the RHS, which concludes the proof.
