Olympiad inequality 
Let $a,b,c,d \in [0,1]$.
  Prove 
  $$ \frac {a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}+abcd\le3$$

Hi. Im a school student so ive been unable to look for ways to apply this problem. I attempted to use AM-GN on 1+a etc but the bound $2\sqrt{a}$ was too large and was easily larger than 3. Thus id like some help solving this problem.

We need to prove that
$$\sum_{cyc}\left(\frac{a}{1+b}-a\right)+abcd\leq3-a-b-c-d$$ or
$$3-a-b-c-d+\sum_{cyc}\frac{ab}{1+b}\geq abcd.$$
Now, by C-S and the given condition $$\sum_{cyc}\frac{ab}{1+b}=\sum_{cyc}\frac{abcd}{cd+bcd}\geq\frac{16abcd}{\sum\limits_{cyc}(cd+bcd)}\geq\frac{16abcd}{8}=2abcd.$$
Thus, it's enough to prove that
$$3-a-b-c-d+abcd\geq0,$$ which is a linear inequality of all variables.
Thus, $$\min_{\{a,b,c,d\}\subset[0,1]}(3-a-b-c-d+abcd)=\min_{\{a,b,c,d\}\subset\{0,1\}}(3-a-b-c-d+abcd)=0$$ and we are done!
