How prove this $\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+e}+\frac{d-e}{d+2e+a}+\frac{e-a}{e+2a+b}\ge 0$ Let $a,b,c,d,e$ are postive real numbers,show that
$$\dfrac{a-b}{a+2b+c}+\dfrac{b-c}{b+2c+d}+\dfrac{c-d}{c+2d+e}+\dfrac{d-e}{d+2e+a}+\dfrac{e-a}{e+2a+b}\ge 0$$
My try: since
$$\Longleftrightarrow\sum_{sym}\left(\dfrac{a-b}{a+2b+c}+\dfrac{1}{2}\right)\ge\dfrac{5}{2}$$
$$\Longleftrightarrow \sum_{sym}\left(\dfrac{3a+c}{a+2b+c}\right)\ge 5$$
use Cauchy-Schwarz inequality,we have
$$\sum_{sym}\dfrac{3a+c}{a+2b+c}\sum_{sym}((3a+c)(a+2b+c))\ge\left(\sum_{sym}(3a+c)\right)^2$$
$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$
becasue this is not hold 
$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$for $a,b,c,d,e>0$
and this method  is from this simaler inequality :
How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$
then I can't,Thank you

You need to try a different fraction to add, so that you get better terms on applying Cauchy-Schwarz.  For e.g. here we have the equivalent inequality:
$$\sum_{cyc} \left(\frac{a-b}{a+2b+c}+\frac15\right)\ge 1$$  
$$\iff \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge 5$$  
By Cauchy-Schwarz, we have:
$$ \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge \frac{\left(\sum_{cyc} (6a-3b+c)\right)^2}{\sum_{cyc}\left((a+2b+c)(6a-3b+c)\right)} = \frac{16(a+b+c+d+e)^2}{\sum_{cyc}(a^2+8ab+7ac)}$$
So it is sufficient if we can show that:
$$16(a+b+c+d+e)^2 \ge 5 \sum_{cyc}(a^2+8ab+7ac) $$
But we can express the above $LHS - RHS$ as:
$$4\sum_{cyc}(a-b)^2 + \frac32 \sum_{cyc}(a-c)^2 \ge 0$$

Added based on a comment in linked post - this application of Cauchy Schwarz requires the numerator of the fraction $6a-3b+c$ to be non-negative, and hence covers only the cases with a condition like $a,b,c,d,e \in [\frac1k,k]$ where $k^2=\frac73$. 
