BMO 2022 shortlist

Let $S=a+b+c+d$. By AM-HM (or Cauchy-Schwarz) we have $$S+4=(a+1)+(b+1)+(c+1)+(d+1)\ge \frac{16}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}}=\frac{16}{3}$$ giving $S\ge \frac{4}{3}$. Multiplying the given equality by $(a+1)(b+1)(c+1)(d+1)$ we get $$\sum abc+2\sum ab+3S+4=3\left(abcd+\sum abc+\sum ab+S+1\right)$$ giving $$3abcd+2\sum abc+\sum ab=1.$$ In particular $ab+ac+ad+bc+bd+cd\le 1$. So we may assume that $S<2$ as otherwise the inequality is immediate. The given equality transforms to $$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}=1,$$ and so by Cauchy-Schwarz $$\sum a(a+1)\sum \frac{a}{a+1}\ge S^2.$$ Thus $$S^2\le a^2+b^2+c^2+d^2+S=S^2-2\sum ab+S.$$ So $\sum ab\le S/2$ and it is enough to prove that $$\frac{3S}{2}+\frac{4}{S}\le 5.$$ This is equivalent to $3S^2-10S+8\le 0$ which in turn is equivalent to $(S-2)(3S-4)\le 0$. Since $S\ge 4/3$ and we are also assuming that $S<2$, then the inequality is true and the result follows.