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?
 A: 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 $f(0) \geqslant f(2)$
A: 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.
A: 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 $
A: I have an alternative proof.
We need to prove that
$$
\frac{a}{b} + 
\frac{b}{c} + 
\frac{c}{a} \geqslant 
\frac{a + 2}{b + 2} + 
\frac{b + 2}{c + 2} + 
\frac{c + 2}{a + 2}
$$
Here we can write this inequality in $2$ forms:
$$
\frac{c}{a} - 
\frac{c + 2}{a + 2} \geqslant
\frac{a + 2}{b + 2} - 
\frac{a}{b} +
\frac{b + 2}{c + 2} - 
\frac{b}{c} 
$$
$$
\frac{c - a}{a^2 + 2a} = 
\frac{b - a}{a^2 + 2a} + \frac{c - b}{a^2 + 2a}
\geqslant 
\frac{b - a}{b^2 + 2b} + 
\frac{c - b}{c^2 + 2c}
$$
And
$$
\frac{b}{c} - \frac{b + 2}{c + 2} + 
\frac{c}{a} - \frac{c + 2}{a + 2} \geqslant
\frac{a + 2}{b + 2} - \frac{a}{b}
$$
$$
\frac{b - c}{c^2 + 2c} + 
\frac{c - a}{a^2 + 2a} \geqslant
\frac{b - a}{b^2 - 2b}
$$
Let $\min{(a,b,c)} = a$.
Case I: $c\geqslant b\geqslant a$: Write inequality in the first form.
Case II: $b\geqslant c\geqslant a$: Write inequality in the second form.
