THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

We notice that $$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}+abcd\le \frac{a}{1+abcd}+\frac{b}{1+abcd}+\frac{c}{1+abcd}+\frac{d}{1+abcd}+abcd=\frac{a+b+c+d}{1+abcd}+abcd.$$ Using repeatedly the inequality $x+y\le 1+xy$, $\forall x,y\in [0,1]$ (equivalent to $(1-x)(1-y)\ge 0$, therefore true), we get $a+b+c+d\le 1+ab+1+cd=ab+cd+2\le 1+abcd+2=abcd+3$.

Replacing $x=abcd\in [0,1]$, it is enough to prove that $1+\frac{2}{1+x}+x\le 3$, that is $2+x^2+x\le 2x+2$, or $x(1-x)\ge 0$, which is clearly true.