Inequality with five variables
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.
A proof for $n=3$.
We'll prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$ for all positives $a$, $b$ and $c$.
Indeed, let $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}$.
Hence, by C-S $\sum\limits_{cyc}\frac{a}{a+b}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}=\frac{1}{1-\frac{ab+ac+bc}{(a+b+c)^2}}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$.
Let $ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}$.
Hence, by C-S $\sum\limits_{cyc}\frac{a}{a+b}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2c^2+a^2bc)}=\frac{1}{1-\frac{abc(a+b+c)}{(ab+ac+bc)^2}}\geq\frac{1}{1-\frac{\sqrt[3]{abc}}{a+b+c}}=\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$.
Done!