<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 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.
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