<s>[INST] How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$ let $a,b,c>0$,and such 
$a+b+c=3$,
show that
$$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$
I think this inequality use this
$$ab\le\dfrac{(a+b)^2}{4}$$
 [/INST] By C-S $$\sum_{cyc}\frac{1}{(a+b)(4-ab)}=\sum_{cyc}\frac{(c+6)^2}{(a+b)(4-ab)(c+6)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(c+6)\right)^2}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}=\frac{441}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}.$$
Thus, it remains to prove that
$$882\geq\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, it's obvious that the last inequality is a linear inequality of $w^3$,
which says that it's enough to prove the last inequality for the extremal value of $w^3$,
which happens in te following cases.


*

*$w^3\rightarrow0^+$.


Let $c\rightarrow0^+$ and $b=3-a$.
After this substitution we obtain:
$$882-\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2=18>0;$$


*$b=a$ and $c=3-2a$.


In this case we get $(a-1)^2(2a^3-3a^2+6a+3)\geq0$, which is obvious.
Done!
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