<s>[INST] prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$ 
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$

My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.
The case $a\ge c\ge b$ is analogous.
Thus Proved!
Is it correct?...And any other alternative ways possible?
 [/INST] I apologize for the first non-obvious proof and I give you a refinement
Hint :$a\geq b \geq c$
First prove that :
$$\frac{2(x-y)}{x+y+y^2}\leq \frac{x}{y}-\frac{x+2}{y+2}\quad (1)$$
For that multiply by $y(y+2)(x+y+y^2)$ , put in factor and it becomes :
$$2(x-y)^2\geq 0$$
Apply $(1)$ for $(a,b)$,$(b,c)$,$(c,a)$
Now we need to show :
$$\frac{2(a-b)}{a+b+b^2}+\frac{2(b-c)}{b+c+c^2}+\frac{2(c-a)}{a+c+a^2}\geq 0$$
Now introducing $f(c)$
$$\frac{2(b-c)}{b+c+c^2}+\frac{2(c-a)}{a+c+a^2}=f(c)$$
Using derivatives prove that $f(c)$ is decreasing when $c$ increases .
Now we put $b=c$ and the inequality becomes :
$$\frac{2(a-b)}{a+b+b^2}+\frac{2(b-a)}{a+b+a^2}\geq 0$$
Wich is obvious with the condition $a\geq b \geq c $
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