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?

Your application of rearrangement is correct, in either case, $(a, b, c)$ and $(a^2+a, b^2+b, c^2+c)$ are similarly ordered, so
$$\sum_{cyc} \frac{a}{a^2+2a} \leqslant \sum_{cyc} \frac{a}{b^2+2b}$$

For another way, which generalises, consider
$$f(x) = \sum_{cyc} \frac{a+x}{b+x}, \quad f'(x) = \sum_{cyc} \frac{b-a}{(b+x)^2} = \sum_{cyc} \frac{b}{(b+x)^2} - \sum_{cyc}\frac{a}{(b+x)^2} \leqslant  0$$
again by Rearrangement.  Hence $f$ is decreasing, and $f(0) \geqslant f(2)$
