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Hello during a problem I have to solve this with $a,b,c$ positiv real number: $$6\geq\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}\geq 3$$ Edit:Wolfram alpha says that the minimum value is 3.

Second edit :This is what I want to solve :

$$\sum_{cyc}\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}\leq \sum_{cyc}\frac{a + \sqrt{a b} + 2 \sqrt{a} + b + 3}{a + \sqrt{a b}- 2 \sqrt{a} + b + 3}$$

I try many classical inequalities , but without success .

Thanks a lot .

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    $\begingroup$ Hint: for all three terms, the denominator is positive and the numerator is greater than it. this give you an lower bound. Can you find a way to approach this lower bound? $\endgroup$ – achille hui Oct 22 '17 at 18:02
  • $\begingroup$ Yes I understand many thanks ! $\endgroup$ – user448747 Oct 22 '17 at 18:11
  • $\begingroup$ @achille hui Very nice solution! $\endgroup$ – Michael Rozenberg Oct 22 '17 at 18:15
  • $\begingroup$ @FatsWallers I solved your problem. Why did you change the problem? You can open another topic for this. I think it's not fair! $\endgroup$ – Michael Rozenberg Oct 29 '17 at 16:41
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    $\begingroup$ Please refrain from changing the question once an answer has been provided. Chamaleon questions should be avoidedat all costs, since they tend to waste the efforts (and time) of users. Please use separate questions for different separate things. $\endgroup$ – Jack D'Aurizio Oct 29 '17 at 17:03
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Your first inequality is wrong take $a=\frac{1}{1000}$ and $a=b=1.5$

It's more than 6 .

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The inequality

$$\sum_{cyc}\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}\leq \sum_{cyc}\frac{a + \sqrt{a b} + 2 \sqrt{a} + b + 3}{a + \sqrt{a b}- 2 \sqrt{a} + b + 3}.$$

is wrong.

Try $b=c\rightarrow0^+$ and $a=1.1$.

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