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

Let $$a,b,c,d&gt;0,a+b+c+d=4$$
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 $a,b,c&gt;0,a+b+c=3$, 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 $$a,b,c,d&gt;0,a+b+c+d=4$$
show that
$$P=\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$

Now using Cauchy Schwarz:
$$\left(\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\right)((a+bc)+(b+cd)+(c+da)+(d+ab))\ge (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2$$
Now:
$$P\ge\frac{4+2(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd})}{4+ab+bc+cd+da}$$
Let $S_1=\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd},S_2=ab+bc+cd+da$
Now we need to prove:
$$\frac{4+2S_1}{4+S_2}\ge2\implies S_1-S_2\ge2$$
If we prove $S_1\ge6$ and $S_2\le4$, then $S_1-S_2\ge6-4=2$. Both inequalities can be proved easily and both hold when $a=b=c=d=1$:
$$(a+b+c+d)(b+c+d+a)\ge(ab+bc+cd+da)^2\implies S_2^2\le16\\\text{ since }a,b,c,d&gt;0\implies S_2\le4$$
Take $1^{st}$ and $3^{rd}$ term apply the inequality, then take $2^{nd}$ and $4^{th}$ term and apply the inequality, you get 

$$\ge 2\left(\frac1{1+\sqrt{bd}}+\frac1{1+\sqrt{ac}}\right)$$

again apply the inequality you get $\displaystyle \ge 2\left(\frac2{1+\sqrt{abcd}}\right)$

for $abcd =1$ you get your equality ie $a=b=c=d=1$