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

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
