75th Romanian Mathematical Olympiad

Adding $4$ to both sides of the second equation, we write: $$1+a+\frac{1+a+b}{1+a}+\frac{1+a+b+c}{1+a+b}+\frac{1+a+b+c+d}{1+a+b+c}=12.$$ Applying the AM-GM inequality successively, we obtain: $$\begin{array}{rl}1+a+\frac{1+a+b}{1+a}& \ge 2\sqrt{(1+a)\cdot \frac{1+a+b}{1+a}}=2\sqrt{1+a+b};\\ \frac{1+a+b+c}{1+a+b}+\frac{1+a+b+c+d}{1+a+b+c}& \ge 2\sqrt{\frac{1+a+b+c}{1+a+b}\cdot \frac{1+a+b+c+d}{1+a+b+c}}=\\ & \ge 2\sqrt{\frac{81}{1+a+b}}.\end{array}$$ By adding these two inequalities and applying the AM-GM inequality again, we have: $$\begin{array}{rl}12& =1+a+\frac{1+a+b}{1+a}+\frac{1+a+b+c}{1+a+b}+\frac{1+a+b+c+d}{1+a+b+c}\\ & \ge 2\sqrt{1+a+b}+\frac{18}{\sqrt{1+a+b}}\ge 12.\end{array}$$

From this, we obtain $a+b=8$ and $$1+a=\frac{1+a+b}{1+a}=\frac{1+a+b+c}{1+a+b}=\frac{1+a+b+c+d}{1+a+b+c}=3.$$ Therefore, the desired numbers are $a=2$, $b=6$, $c=18$, $d=54$.