Prove, inequality ,positive numbers $$\frac{a}{e+a+b}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{c+d+e}+\frac{e}{d+e+a}<2$$
Prove that for positive numbers $a,b,c,d,e$ there is such inequality
 A: For convenience of notation I'll use $a_i$'s ($1 \le i \le 5$) instead of $a,b,c,d,e$.
Wlog assume $a_1+a_2+a_3 = \min\limits_{1 \le i \le 5}\{a_{i-1}+a_i+a_{i+1}\}$ 
(where, the indexing is cyclic $a_{0} = a_{5}$ and $a_6 = a_1$ and so on)
Then, $\displaystyle \sum\limits_{i=1}^{3} \frac{a_i}{a_{i-1}+a_i+a_{i+1}} \le \frac{a_1}{a_1+a_2+a_3}+\frac{a_2}{a_1+a_2+a_3}+\frac{a_3}{a_1+a_2+a_3} = 1$
And, $\displaystyle \frac{a_4}{a_3+a_4+a_5}+\frac{a_5}{a_4+a_5+a_1} < \frac{a_4}{a_4+a_5} + \frac{a_5}{a_4+a_5} = 1$.
Adding together we get:
$$\displaystyle \sum\limits_{i=1}^{5} \frac{a_i}{a_{i-1}+a_i+a_{i+1}} < 2$$
Infact using a $5$-tuple, $(a_1,a_2,a_3,a_4,a_5) = (x,y,x,y,x)$ and letting $y \to \infty$,
$$\lim\limits_{y \to \infty} \sum\limits_{i=1}^{5} \frac{a_i}{a_{i-1}+a_i+a_{i+1}} =\lim\limits_{y \to \infty} \frac{3x}{x+2y}+\frac{2y}{2x+y} = 2$$
Hence, $2$ is the best possible upper-bound here.
A: We can write our equation as $\frac{a}{s}+\frac{b}{g}+\frac{c}{r}+\frac{d}{p}+\frac{e}{x}<2$.
 After some operations we get to $s(x(p(br+cg)+d(gr))+egrp)$
and then
$sgrpx(\frac{p(br+cg)+dgr}{grp}-\frac{2grp}{grp})<-agrpx-se$
and again
$s(\frac{pbr+pcg+dgr-2grp}{grp})<-a-\frac{e}{grpx}$
so
$\frac{pbr+pcg+dgr-2grp}{grp}<0$
and 
$-2grp<0$
$-2(a+b+c)(b+c+d)(c+d+e)<0$
