Prove that $\sum\limits_{\mathrm{cyc}} \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3} \le 6$ for $a, b, c \ge 1$ 
I have already post this inequality but I add this condition $a,b,c\geq 1$
$$6\geq \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}$$

I try many classical inequalities but without success...
My try :
We study the following function :
$$f(x)=\frac{x^2+x+2x+3+1}{x^2+x-2x+3+1}+\frac{x^2+x+2+3+1}{x^2+x-2+3+1}+2$$
It's decreasing and the maximum is reached for $x=1$ and is equal to 6.
Now we put :
$$x= \frac{(\sqrt{4 a^3 + 4 a^2 b - 15 a^2 + 4 a b^2 + 12 a} - a)}{2 a}$$
We get :
$$f(x)=\frac{(a^2+b^2+ab+3+2a)}{(a^2+b^2+ab+3-2a)}+2+\frac{(a^2 + \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + ab-a + b^2 + 3)}{(a^2 - \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + a b + a + b^2 + 3)} $$
So if $a\geq b \geq c$ and if we suppose:$\quad$ $max(\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3};\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3})=\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}$ it remains to prove that we have :
$$\frac{(a^2 + \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + ab-a + b^2 + 3)}{(a^2 - \sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)} + a b + a + b^2 + 3)}\geq \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}$$
Wich is true for $b\geq2$ because we have with this condition :
$$\sqrt{a (4 a b + a (4 a - 15) + 4 b^2 + 12)}-a\geq 2a$$
Edit : I have the end of my proof but it's ugly...Now we know that the inequality is true .
Have you easier ?
Thanks a lot .

The given inequality can be transformed into 
$$\sum \frac {a^2+b^2+ab+2a+3}{a^2+b^2+ab-2a+3}\le 6$$
Which changes to 
$$\sum 1+\frac {4a}{a^2+b^2+ab-2a+3}\le 6$$
$$\Rightarrow \sum  \frac {4a}{a^2+b^2+ab-2a+3}\le 3$$
Hence it suffices to show that 
$$\sum \frac {a}{a^2+b^2+ab-2a+3}\le \frac{3}{4}$$
Hence by AM GM we have LHS as 
$$\sum \frac {a}{3(ab+1)-2a}$$
Again by AM GM LHS becomes
$$\sum \frac {a}{6\sqrt {ab}-2a}=\sum \frac {1}{6\sqrt {\frac {b}{a}}-2}$$ 
Hence it suffices to show that 
$$\sum \frac {1}{3\sqrt {\frac {b}{a}}-1}\le \frac {3}{2}$$ 
That means 
$$\frac {1}{3\sqrt {\frac {b}{a}}-1}+    \frac {1}{3\sqrt {\frac {c}{b}}-1}+   \frac {1}{3\sqrt {\frac {c}{a}}-1}\le \frac {3}{2}$$
I hope you can take it from here. 
