it's just a clarification about a proof :
The starting inequality is :
$$0.5\geq1.5-\frac{1}{\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+1}-\frac{1}{\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+1}-\frac{1}{\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}+1}$$
So we study the function $f(x)=0.5-\frac{1}{x+1}$ wich is concave so we can apply Jensen inequality we get : $$0.5-\frac{1}{\frac{\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+{\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}}}{3}+1}\geq\frac{1}{3}(1.5-\frac{1}{\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+1}-\frac{1}{\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+1}-\frac{1}{\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}+1})$$
Wich is equivalent to :
$$1.5-\frac{3}{\frac{\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+{\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}}}{3}+1}\geq(1.5-\frac{1}{\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+1}-\frac{1}{\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+1}-\frac{1}{\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}+1})$$
But it's easy to find the minimum of :
$$\frac{3}{\frac{\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+{\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}}}{3}+1}$$
Wich is one so we have :
$$0.5\geq 1.5-\frac{1}{\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+1}-\frac{1}{\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+1}-\frac{1}{\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}+1}$$
Done !
But Michael Rozenberg says that there is a mistake but I don't see where .
Edit : It's related to this
Thanks a lot.
You proved that $$\sum_{cyc}\frac{1}{\frac{a^2+ab+2a+b^2+3}{a^2+ab-2a+b^2+3}+1}\geq\frac{3}{\frac{\sum\limits_{cyc}\frac{a^2+ab+2a+b^2+3}{a^2+ab-2a+b^2+3}}{3}+1}$$ or $$\sum_{cyc}\frac{a^2+ab-2a+b^2+3}{a^2+ab+b^2+3}\geq\frac{18}{\sum\limits_{cyc}\frac{a^2+ab+2a+b^2+3}{a^2+ab-2a+b^2+3}+3}.$$
Now, you want to use the following reasoning.
Since $$\sum\limits_{cyc}\frac{a^2+ab+2a+b^2+3}{a^2+ab-2a+b^2+3}=3+\sum\limits_{cyc}\left(\frac{a^2+ab+2a+b^2+3}{a^2+ab-2a+b^2+3}-1\right)=$$ $$=3+\sum\limits_{cyc}\frac{4a}{a^2+ab-2a+b^2+3}\geq3,$$ where the equality occurs for $a=b=c=0$, we see that the minimal value of $$\sum\limits_{cyc}\frac{a^2+ab+2a+b^2+3}{a^2+ab-2a+b^2+3}$$ is $3$.
Thus, it's enough to prove that $$\sum_{cyc}\frac{a^2+ab-2a+b^2+3}{a^2+ab+b^2+3}\geq\frac{18}{3+3}$$ or $$\sum_{cyc}\frac{a^2+ab-2a+b^2+3}{a^2+ab+b^2+3}\geq3$$ and here you say that the starting inequality is proven.
I think it's not so because the last inequality is wrong.
Try $a=b=c=1$.
I hope it's clear now.