How to prove inequality $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ Question:

Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$

when I solved this problem, I have see following three variables inequality:
Assumming that $a,b,c>0,a+b+c=3$, show that :$$f(a,b,c)=\dfrac{a}{a+bc}+\dfrac{b}{b+ca}+\dfrac{c}{c+ab}\ge\dfrac{3}{2}$$
solution can see:inequality
I found this three answer all is not true,
1、such as  dear @Macavity, in fact
$$\sum_{cyc}\dfrac{a^2}{a^2+abc}\ge\dfrac{16}{\sum_{cyc}(a^2+abc)}$$
the Right not
$\dfrac{16}{4+\sum_{cyc}abc}$
2、and the @ante.ceperic is also not true.in fact
$$a^2+b^2+c^2+d^2+abc+bcd+cda+dab\le 8$$ is not true with $a+b+c+d=4$
such let $a=3$

Let's denote the left-hand expression by L.
Then, by Cauchy-Schwarz:
$[a(a + bc) + b(b + cd) + c(c + da) + d(d+ab)]L \geq (a + b + c + d)^2 = 16$
Now, we have to show 
$[a(a + bc) + b(b + cd) + c(c + da) + d(d+ab)] \leq 8$ and we are over.
Let's write it down like:
$A(a, b, c, d) = a^2 + b^2 + c^2 + d^2 + ac(b+d) + bd(a+c) \leq 8$
Notice that we get equality for $a = c, b = d$. I'll try this tactic:
if I prove that $A(a, b, c, d) \leq A(\frac{a + c}{2}, b, \frac{a+c}{2}, d)$ and $A(a, b, c, d) \leq A(a, \frac{b + d}{2}, c, \frac{b + d}{2})$ (these two statements are analogous, so I'll just try to show the first) we will have this chain:
$A(a, b, c, d) \leq A(\frac{a + c}{2}, b, \frac{a+c}{2}, d) \leq A(\frac{a + c}{2}, \frac{b + d}{2}, \frac{a+c}{2}, \frac{b + d}{2}) = 8$.
Let's evaluate:
$A(\frac{a + c}{2}, b, \frac{a+c}{2}, d) - A(a, b, c, d) =
\\ = 2(\frac{a + c}{2})^2 - a^2 - c^2 +(b+d)[(\frac{a + c}{2})^2 - ac]
\\ = -2(\frac{a - c}{2})^2 + [4 - (a+c)](\frac{a - c}{2})^2
\\ = (\frac{a - c}{2})^2[2 - (a+c)]$
This works just if $a + c \leq 2$. If it's not true, $b + d \leq 2$ is, so we can start with the other substitution, and make at least one step in our chain. So, we can assume $a + c \leq 2$ and now we deal with:
$2a^2 + b^2 + d^2 + a^2(b + d) + 2abd \leq 8$
where $2a + b + d = 4, a \leq 1$.
In the next few calculations, I'm just using AM-GM on $bd$ and $b + d = 4 - 2a$.
$2a^2 + b^2 + d^2 + a^2(b + d) + 2abd 
\\ = 2a^2 + (b+d)^2 + a^2(b + d) + 2(a-1)bd
\\ = 2a^2 + 4(2-a)^2 + 2a^2(2 - a) + 2(a - 1)bd
\\ \leq 2a^2 + 4(2-a)^2 + 2a^2(2 - a) + 2(a - 1)(2-a)^2$
After tidying this up a little bit, it is easy to see that for $a = 0$ and $a = 1$ upper expression equals 8. Differentiate it twice to show that it's convex, and we are finished.
