, 2022

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.