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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?

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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 $f(0) \geqslant f(2)$

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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.

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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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  • $\begingroup$ nice(+1) but i am getting something different in your second step,are you sure $\endgroup$ Nov 1, 2020 at 2:48
  • $\begingroup$ Oh yeah! The proof is faulty, I am sorry. $\endgroup$ Nov 1, 2020 at 2:58
  • $\begingroup$ @AlbusDumbledore I have corrected it. Please take a look. $\endgroup$ Nov 1, 2020 at 5:49
  • $\begingroup$ very neat! thanks! $\endgroup$ Nov 1, 2020 at 6:53
  • $\begingroup$ Actually its a bit similar to the proof of rearrangement inequality,still it is good $\endgroup$ Nov 1, 2020 at 6:59
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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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  • $\begingroup$ @Erik Satie How you want to use SOS here? Explain please. $\endgroup$ Oct 31, 2020 at 18:24

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