<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] Your proof is nice and right.
Another way.
Let $c=\min\{a,b,c\}$.
Thus, we need to prove that:
$$\frac{a}{b}+\frac{b}{a}-2+\frac{c}{a}-\frac{b}{a}+\frac{b}{c}-1\geq\frac{a+2}{b+2}+\frac{b+2}{a+2}-2+\frac{c+2}{a+2}-\frac{b+2}{a+2}+\frac{b+2}{c+2}-1$$ or
$$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq\frac{(a-b)^2}{(a+2)(b+2)}+\frac{(c-a)(c-b)}{(a+2)(c+2)},$$ which is obvious.
</s>