Inequality with five variables Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$
Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is true
and for even $n$ and positives $a_i$ the following inequality is true.
$$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$

Here is a full proof.
Let us start the discussion for general  $n$. Denote $S = \sum_{i=1}^n a_i$.
Since by AM-GM, $S \geq n \sqrt[n]{a_1a_2...a_n}$, we have
$$1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} \geq \frac{S}{S - (n-2)\sqrt[n]{a_1a_2...a_n}}$$
Hence a  tighter claim is (simultaneously defining $L$ and $R$):
$$L = \sum_{cyc}\frac{a_i}{a_i+a_{i+1}}\geq 1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} = R$$
and it suffices to show that one.
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 our final equivalent reformulation of the L-R claim above, 
$$ \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} \leq (n -  2) (1- \frac{n \sqrt[n]{a_1a_2...a_n}}{S} )$$
For general odd $n$ see the remark at the bottom. Here 
the task is to  show $n=5$.  
Before doing so, we will first prove the following Lemma (required below), which is the above L-R-inequality for 3 variables (which is tighter than the original formulation, hence we cannot apply the proof for $n=3$ given above  by Michael Rozenberg for the original formulation):
$$  \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c} \leq  (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )$$
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 prove, under that restriction,   $$  \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq  1+\frac{3}{2(a+b+c)}$$
This reformulates into 
 $$ \frac{a\; c}{a +b} + \frac{b\; a}{b +c} + \frac{c\; b}{c +a} \geq \frac{3}{2}$$
or equivalently, due to $abc=1$,
$$ \frac{1}{b(a +b)} + \frac{1}{c(b +c)} + \frac{1}{a(c +a)} \geq \frac{3}{2}$$ 
which is known (2008 International Zhautykov Olympiad), for some proofs see here: http://artofproblemsolving.com/community/c6h183916p1010959 
Hence the Lemma holds.
For $n=5$, we rewrite the LHS of our above final reformulation by adding and subtracting terms:
$$ \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{d-c}{d+c} + \frac{e-d}{e+d} + \frac{a-e}{a+e} = \\
(\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + (\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e})
$$ 
This also holds for any cyclic shift in (abcde), so we can write
$$ 5 (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{d-c}{d+c} + \frac{e-d}{e+d} + \frac{a-e}{a+e}) = \\
\sum_{cyc (abcde)} (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + \sum_{cyc (abcde)}(\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + \sum_{cyc (abcde)} (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e})
$$ 
Using our Lemma, it suffices to show (with $S = a +b+c+d+e$)
$$
\sum_{cyc (abcde)} (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )
 + \sum_{cyc (abcde)}(1- \frac{3 \sqrt[3]{a\, c \, d}}{a + c+ d} )
+ \sum_{cyc (abcde)}(1- \frac{3 \sqrt[3]{a\, d \, e}}{a + d+ e} )
\leq 15 (1- \frac{5 \sqrt[5]{a b c d e }}{S} )
$$
which is
$$
\sum_{cyc (abcde)} (\frac{\sqrt[3]{a\, b \, c}}{a + b+ c} + \frac{\sqrt[3]{a\, c \, d}}{a + c+ d} + \frac{\sqrt[3]{a\, d \, e}}{a + d+ e} )
\geq 25  \frac{\sqrt[5]{a b c d e }}{S}
$$
Using Cauchy-Schwarz leaves us with showing
$$
\frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2}{\sum_{cyc (abcde)}(a + b+ c)} 
+
\frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, c \, d})^2}{\sum_{cyc (abcde)}(a + c+ d)} 
+
\frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, d \, e})^2}{\sum_{cyc (abcde)}(a + d+ e)} 
\geq 25  \frac{\sqrt[5]{a b c d e }}{S}
$$
The denominators all equal $3S$, so this becomes
$$
(\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2 
+
(\sum_{cyc (abcde)} \sqrt[6]{a\, c \, d})^2
+
(\sum_{cyc (abcde)} \sqrt[6]{a\, d \, e})^2 
\geq 75 \sqrt[5]{a b c d e }
$$
Using AM-GM gives for the first term
$$
(\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2 \geq ( 5 (\prod_{cyc (abcde)} \sqrt[6]{a\, b \, c} )^{1/5})^2 
= 25 (\prod_{cyc (abcde)} ({a\, b \, c} ) )^{1/15}  = 25 \sqrt[5]{a b c d e }
$$
By the same procedure, the second and the third term on the LHS are likewise greater or equal than $25 \sqrt[5]{a b c d e }$. This concludes the proof.
Remarks:


*

*the tighter $L-R$-claim used here is - for general $n$ - asked for in the problem given at Cyclic Inequality in n (at least 4) variables 

*For general odd $n$, the above reformulation can be used again. For odd $n>5$, take the method of adding and subtracting terms to form smaller sub-sums which are cyclically closed in a smaller number of variables, and apply previous results for smaller $n$ recursively.  
