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
$$ab\le\dfrac{(a+b)^2}{4}$$

It is possible to prove a slightly weaker bound by cutting down the dimension of the problem.
For any $K\in[0,3]$, define:
$$f_K(x)=\frac{2}{(3-x)(4-x(K-x))},\qquad g_K(x)=f_{K}(x)+f_{K}(K-x).$$
By differentiating with respect to $x$, we have that the minima of $g_K(x)$ over $[0,3]$ are located in $\frac{K}{2}\pm\frac{1}{2}\sqrt{1+(3-K)^2}$, so
$$g_K(x) \geq g_K\left(\frac{K}{2}\pm\frac{1}{2}\sqrt{1+(3-K)^2}\right)=\frac{8(6-K)}{(13-3K)^2}$$
holds over $[0,K]$. By taking $x=b,K=b+c$, $a=3-K$ follows and we have:
$$\sum_{cyc}\frac{2}{(3-a)(4-bc)}=\frac{2}{K(4-x(K-x))}+g_K(x).\tag{1}$$
The first term in the RHS is a non-negative and convex function $h_K(x)$ on $[0,K]$ whose graphics is symmetric with respect to $x=K/2$. Since $h_K(K/2)>h_K(0)$, $h_K(x)\geq h_K(0)=\frac{1}{2K}$ follows. This gives:
$$\sum_{cyc}\frac{2}{(3-a)(4-bc)}\geq\frac{1}{2K}+\frac{8(6-K)}{(13-3K)^2}=j(K).\tag{2}$$
$j(K)$ is a convex function over $[0,3]$, whose minimum is attained in $x=1.4638\ldots$. This gives:
$$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq 0.831262\ldots > \frac{4}{5}.\tag{3}$$
We can improve this to:
$$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq h_K\left(\frac{K}{2}-\frac{1}{2}\sqrt{1+(3-K)^2}\right)+\frac{8(6-K)}{(13-3K)^2}=\frac{4(13+9K-2K^2)}{K(13-3K)^2},\tag{4}$$
where the RHS is increasing on $[2,3]$. Since we can assume $K=(b+c)\geq 2$ without loss of generality, we have:
$$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq\frac{46}{49}.\tag{5}$$
