<s>[INST] If $a+b+c+d=4$, then $\sum\frac{1}{a+3}\le \frac{1}{abcd}$ This is somehow related to this problem but I don't have any idea about it.

Let $a$, $b$, $c$ and $d$ be positive reals such that $a+b+c+d=4$. Prove that:
  $$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\le \frac{1}{abcd}$$

Now I also tried to prove the $3$ variable version :
$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\le \frac{1}{abc}$$ where $a,b,c>0$ with $a+b+c=3$. But I haven't been able to solve it too. Anyone can help?? Thanks a lot.
 [/INST] $\sum$ will be used to denote the cyclic sum.
For the second inequality, let $t=ab+bc+ca$. Note that $t\le 3$, and also:
$$3t=(a+b+c)(ab+bc+ca)\ge 9\sqrt[3]{(abc)(a^2b^2c^2)}=9abc$$
Now, by Cauchy-Schwarz:
$$\sum\frac{2bc}{a+2}=2\sum\frac{(bc)^2}{abc+2bc}\ge \frac{2t^2}{3abc+2t}\ge \frac{2t}{3}$$
Note further that $\frac{2t}{3}-(t-1)=\frac{3-t}{3}\ge 0$ since $t\le 3$. Hence:
$$\sum\frac{2bc}{a+2}\ge ab+bc+ca-1\implies \sum (bc-\frac{2bc}{a+2})\le 1\implies \sum\frac{1}{a+2}\le\frac{1}{abc}$$
as desired. 
A similar trick works for the first inequality. Let $t=abc+bcd+cda+dab$ this time. Then by repeated applications of AM-GM, 
$$t=ab(c+d)+cd(a+b)\le (\frac{a+b}{2})^2(c+d)+(a+b)(\frac{c+d}{2})^2=\frac{(a+b+c+d)^2}{4}=4$$
Also, note that by AM-GM:
$$4t=(a+b+c+d)(abc+bcd+cda+dab)\ge 16\sqrt[4]{(abcd)(a^3b^3c^3d^3)}=16abcd$$
Using these estimations:
$$\sum \frac{3bcd}{a+3}=3\sum\frac{(bcd)^2}{abcd+3bcd}\ge \frac{3t^2}{4abcd+3t}\ge\frac{3t}{4}$$
And $\frac{3t}{4}-(t-1)=\frac{4-t}{4}\ge 0$, so that:
$$\sum\frac{3bcd}{a+3}\ge abc+bcd+cda+dab-1\implies \sum (bcd-\frac{3bcd}{a+3})\le 1\implies \sum\frac{1}{a+3}\le \frac{1}{abcd}$$
as desired. 
Note: Using this same method, the $n$-variable inequality can also be proven. 
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