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 $abcd=1,a,b,c,d&gt;0$

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&amp;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
We have
$$\begin{align}
\frac3{a+b+c+1} &amp;- \frac1{a+3} - \frac1{b+3} - \frac1{c+3}\\
&amp;= \sum_{cyc}^{a, b, c} \left(\frac1{a+b+c+1}-\frac1{a+3} \right) \\
&amp;= \frac1{a+b+c+1} \sum_{cyc}^{a, b, c}\frac{2-b-c}{a+3} \\
&amp;= \frac1{(a+b+c+1)\prod_{cyc}^{a, b, c}(a+3)} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2) \\
&amp;\le \frac1{1\times3^3} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2)
\end{align}$$

using $a, b, c &gt; 0$. 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!
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,
$${\large \sum_{cyc}\frac{1}{a_1+(n-1)}\ge\sum_{cyc,i}\frac1{1+\sum_{cyc,j\ne i}a_j}}\Box$$