<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] Let $a+b+c+d=4u$ and $ab+ac+ad+bc+bd+cd=6v^2$.
Hence, by AM-GM $abcd\leq v^4$ and by C-S
$$\frac{1}{abcd}-\sum_{cyc}\frac{1}{a+3}=\frac{1}{abcd}-\frac{4}{3}-\sum_{cyc}\left(\frac{1}{a+3}-\frac{1}{3}\right)=$$
$$=\frac{1}{abcd}-\frac{4}{3}+\sum_{cyc}\frac{a}{3(a+3)}\geq\frac{1}{abcd}-\frac{4}{3}+\frac{(a+b+c+d)^2}{3\sum\limits_{cyc}(a^2+3)}=$$
$$=\frac{1}{abcd}-\frac{4}{3}+\frac{16}{3\sum\limits_{cyc}(a^2+3)}=\frac{u^2}{abcd}-\frac{4}{3u^2}+\frac{16}{3(16u^2-12v^2+12u^2)}\geq$$
$$\geq\frac{u^2}{v^4}-\frac{4}{3u^2}+\frac{4}{3(7u^2-3v^2)}\geq0,$$
where the last inequality it's 
$$(u^2-v^2)(7u^4+4u^2v^2-4v^4)\geq0,$$
which is true because $u^2\geq v^2$.
Done!
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