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

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 ...
