<s>[INST] 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
 [/INST] I'll solve the three variable case (the $a,b,c,d,e$ case is similar).
Assume without loss of generality $a \geq b \geq c$. Multiplying by the denominators we obtain
$$
a^3+a^2 c+a b^2-6 a b c+b^3+b c^2+c^3 \geq 0
$$
Now use the rearrangment inequality twice. We can split the inequality in two parts:
$$
a^2c+b^2a+c^2b \geq 3abc
$$
and
$$
a^3+b^3+c^3 \geq 3abc
$$
For the first one, use the rearrangment inequality to the sequences $ab, ac, bc$ and $a,b,c$ (which are listed in increasing order).
As for the second one, you can use the extension of the rearrangement inequality (also explained in the link) with three equal sequences $a,b,c$. Sum up the two inequalities and you are done.
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