How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$ show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$
I have show three  variable inequality
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.
Prove that
$$\frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$
also see:can  see 
：http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=243
from  this equality,I have see a nice methods:


I think this Four varible  inequality is also true
First,Thank you Aditya answer,But I read it your solution,it's not true
 A: We have
$$\begin{align}
\frac3{a+b+c+1} &- \frac1{a+3} - \frac1{b+3} - \frac1{c+3}\\
&= \sum_{cyc}^{a, b, c} \left(\frac1{a+b+c+1}-\frac1{a+3} \right) \\
&= \frac1{a+b+c+1} \sum_{cyc}^{a, b, c}\frac{2-b-c}{a+3} \\
&= \frac1{(a+b+c+1)\prod_{cyc}^{a, b, c}(a+3)} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2) \\
&\le \frac1{1\times3^3} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2)
\end{align}$$
using $a, b, c > 0$. Summing over four such inequalities, we get
$$3\sum_{cyc}\frac1{a+b+c+1} - 3\sum_{cyc} \frac1{a+3} \\ \le \frac2{27}\left( 108-\sum_{cyc}\left(9(a+a^2)+4(2ab+bd)+(a^2b+ab^2+a^2c+ac^2+b^2d+bd^2)\right) \right)$$ 
Now by AM-GM and the constraint, we have that  $\sum_{cyc} a^mb^n \ge 4\sqrt[4]{(abcd)^{m+n}}=4$ for all $m, n \ge 0$, so RHS $\le 0$ and we are done. 
P.S. the method looks general, though I wouldn't want to write down the cyclic sums for more variables!
A: Partial Proof:
For general case of n variables, the inequality converts to:

$$\sum_i^n \frac1{1+a_1+a_2+\cdots+a_n-a_i}\le \sum_i^n\frac1{n-1+a_i}$$

Similiar to the given proof we can convert $\frac {a_1}{a_1+(n-1)}$ like this:
$$\frac{a_1}{a_1+(n-1)}=\frac{a_1}{a_1+(n-1)(a_1a_2\cdots a_n)^{1/n}}$$
Dividing numerator and denominator by $a_1^{1/n}$:
$$=\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+(n-1)\left(\frac{a_1a_2\cdots a_n}{a_1}\right)^{1/n}}
$$
Finally using AM-GM gives:
$$\frac{a_2+a_3+\cdots+a_n}{(n-1)}\ge\left(a_2a_3\cdots a_n\right)^{1/(n-1)}$$
Or:
$$(n-1)\left(a_2a_3\cdots a_n\right)\le (a_2+a_3+\cdots+a_n)^{n-1}$$
$$=\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+(n-1)\left(\frac{a_1a_2\cdots a_n}{a_1}\right)^{1/n}}
\ge\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+a_2^{(n-1)/n}+\cdots+a_n^{(n-1)/n}}$$
So, $$\sum_i^n\frac{a_i}{a_i+(n-1)}\ge\sum_i^n\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+a_2^{(n-1)/n}+\cdots+a_n^{(n-1)/n}}=1\tag{i}$$
We have proved what we need for the general case of n-variables, try putting $n=3$.

Since product of all numbers is 1, we can define new fractions as: Let $$\displaystyle a_1:=\frac{x_1}{x_2},a_2:=\frac{x_2}{x_3},\cdots,a_n:=\frac{x_n}{x_1}$$
Notice that in the given proof:
$$\frac{b}{ab+b+1}=\frac{x_2/x_3}{x_1/x_2.x_2/x_3+x_2/x_3+1}=\frac{x_2}{x_1+x_2+x_3 }$$
Now similiar to the given proof we can show that(step unproven):
$$\frac{2}{a_1+(n-1)}-\frac{1}{1+a_2+a_3+\cdots+a_n}-\frac{x_2}{x_!+x_2+\cdots+x_n }\ge0$$
$$\frac{2}{a_1+(n-1)}\ge \frac{1}{1+a_2+a_3+\cdots+a_n}+\frac{x_2}{x_1+x_2+\cdots+x_n }$$
$$\frac1{a_1+(n-1)}+\frac1{a_1+(n-1)}\ge\frac{1}{1+a_2+a_3+\cdots+a_n}+\frac{x_2}{x_1+x_2+\cdots+x_n }$$
Since $\displaystyle \sum_i^n\frac{x_2}{x_1+x_2+\cdots+x_n }=1$
$$\frac1{a_1+(n-1)}+\frac1{a_1+(n-1)}\ge\frac{1}{1+a_2+a_3+\cdots+a_n}+1$$
$$\sum_{cyc}\frac{1}{a_1+(n-1)}-\sum_{cyc,i}\frac1{1+a_2+a_3+\cdots+a_n }\ge1-\sum_{cyc}\frac{1}{a_1+(n-1)}\ge0 $$
The$\ge0$ part, we have proved in (i). 
So,
$${\large \sum_{cyc}\frac{1}{a_1+(n-1)}\ge\sum_{cyc,i}\frac1{1+\sum_{cyc,j\ne i}a_j}}\Box$$
A: Let's use Lagrange Multipliers:
Let use take $f$ to be:
$$f(a,b,c,d)=\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)$$
subject to constraint $abcd=1$:
$$g(a,b,c,d)=abcd=1$$
Here are the equations we need to solve:
$$\frac{-1}{(a+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda bcd\\
\frac{-1}{(b+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda acd\\
\frac{-1}{(c+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda abd\\
\frac{-1}{(d+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda abc\\
abcd=1
$$
Let $\displaystyle Z:=\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}$
$$Z=\frac{\lambda}a+\frac1{(a+3)^2}=\frac{\lambda}b+\frac1{(b+3)^2}=\frac{\lambda}c+\frac1{(c+3)^2}=\frac{\lambda}d+\frac1{(d+3)^2}$$
Since $a=b=c=d$ seems to satisfy the equations from intuition. If you don't believe see this page on W|A, which says that when $ab(a+3)(b+3)\ne0$(which is true as $a,b,c,d>0$) the solutions(solved for two variables) are:
$$a=b,\lambda=-\frac{(a b (a+b+6))}{((a+3)^2 (b+3)^2)}$$ 
Since, now $a=b=c=d(=1)$, the value of $f$ seems to be:
$$f(1,1,1,1)=4\times\frac14-4\times\frac14=0$$ 
Which apparently is its minimum value.

Now for the general case of n-variables the equations would be:
$$\frac{-1}{(x_1+(n-1))^2}+\sum_{cyc}\dfrac{1}{(\sum_{cyc}x_i+1)^2}=\lambda \frac{\prod x_i}{x_1}=\frac{\lambda}x_1\tag{cycle equation 'n' times}\\\prod x_i=1$$
Again let $\displaystyle Z:=\sum_{cyc}\dfrac{1}{(\sum_{cyc}x_i+1)^2} $
Now $$Z=\frac{\lambda}x_i+\frac{1}{(x_1+(n-1))^2}$$
Again extending that logic we would get $$x_1=x_2=\cdots=x_n(=1)$$
So the value of $f$ here would be:
$$f(1,1,\cdots,1)=\sum_{i=1}^n\frac{1}{1+\sum_{j=1,j\ne i}^n1}-\sum_{i=1}^n\frac{1}{(n-1)+1}\\=n\times\frac1{1+(n-1)}-n\times\frac1{(n-1)+1}=0$$
Which again is ...
