<s>[INST] 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
 [/INST] 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!
</s>