How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$ let $a,b,c,d,e$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0$$
My try:
I have solved follow Four-variable inequality:
let $a,b,c,d$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge 0$$
poof:since
\begin{align*}
&\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}=\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}-4\\
&=(a+c)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)+(b+d)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)-4
\end{align*}
By Cauchy-Schwarz inequality we have
$$\dfrac{1}{b+c}+\dfrac{1}{d+a}\ge\dfrac{4}{(b+c)+(d+a)},\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{(c+d)+(a+b)}$$
so we get
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge\dfrac{4(a+c)}{(b+c)+(d+a)}+\dfrac{4(b+d)}{(c+d)+(a+b)}-4=0$$
Equality holds for $a=c$ and $b=d$
By done!
But for  Five-variable inequality,I can't prove it.Thank you
 A: For $n$ variables $x_i > 0$, (with the convention $x_{n+k}=x_k$) we need to show that
$$\sum_i \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} \geqslant 0 \iff \sum_i \left( \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} +\frac12 \right) \geqslant \frac{n}2 \iff \sum_i \dfrac{2x_i - x_{i+1}+x_{i+2}}{x_{i+1} + x_{i+2}}  \geqslant n $$
By Cauchy-Schwarz we have
$$\sum_i \dfrac{2x_i - x_{i+1}+x_{i+2}}{x_{i+1} + x_{i+2}} \geqslant   \dfrac{ \left(\sum_i (2x_i - x_{i+1}+x_{i+2}) \right)^2}{\sum_i \left( (x_{i+1} + x_{i+2}) (2x_i - x_{i+1}+x_{i+2} ) \right)} = \dfrac{ 4\left(\sum_i x_i \right)^2}{2\sum_i (x_i x_{i+1} + x_i x_{i+2})} $$
So it is sufficient to show that 
$$ \left(\sum_i x_i \right)^2 \geqslant \dfrac{n}2 \sum_i x_i (x_{i+1}+x_{i+2}) \tag{*}$$
While the general case eludes me, this helps to answer your question and the case $n=6$.
For $n=5$ we have the SOS form:
$$\left(\sum_{i=1}^5 x_i \right)^2 - \dfrac{5}2 \sum_{i=1}^5 x_i (x_{i+1}+x_{i+2}) = \frac14 \sum_{i=1}^5 (x_i - x_{i+1})^2 + \frac14 \sum_{i=1}^5 (x_i - x_{i+2})^2 \geqslant 0$$
For equality we will need all $x_i$ to be the same.
For $n=6$, we can use $a=x_1 + x_4, b = x_2 + x_5, c = x_3+x_6$ to get:
$$\left(\sum_{i=1}^6 x_i \right)^2 - \dfrac{6}2 \sum_{i=1}^6 x_i (x_{i+1}+x_{i+2}) = (a+b+c)^2 - 3(ab+bc+ca) \geqslant 0$$
In this case for equality we need all $x_i$ for odd $i$ to be the same, and similarly all the $x_i$ for even $i$ must also be the same.

Added based on Dongryul Kim's comment - this application of Cauchy Schwarz requires an additional condition like $x_i \in [\frac1{\sqrt3}, \sqrt3]$ so that the numerators in the fraction on LHS are non-negative. 
A: I would have written this in a comment, but I don't have enough 'reputation' to do so...
@Macavity I'd just like to point out that $$\sum \frac{a_i}{b_i} \ge \frac{\big(\sum a_i \big)^2}{\sum a_i b_i}$$ holds when $a_i b_i$ is positive for all $i$. (To use Cauchy-Schwarz, you need to square-root and then square those terms) A counterexample for this might be $a_1=-1, a_2=1, b_1=2$, and $b_2=4$.
Also, this is a conjecture proposed by Vasile Cirtoaje many years ago in Old and New Inequalities (page 60). Inequalities with more variables were discussed here
