NMO Selection Tests for the Junior Balkan Mathematical Olympiad

First solution. We notice the equality case $a = 3$, $b = c = d = 0$ and we try to mix the variables as follows:
$$f(a, b, c, d) \ge f(a + b + c + d, 0, 0, 0) = a + b + c + d - \frac{(a + b + c + d)^2}{3} = 0,$$
where
$$f(a, b, c, d) = \frac{a}{1 + 2b^3} + \frac{b}{1 + 2c^3} + \frac{c}{1 + 2d^3} + \frac{d}{1 + 2a^3} - \frac{a^2 + b^2 + c^2 + d^2}{3}.$$
Then
$$f(a, b, c, d) - a - b - c - d + \frac{(a + b + c + d)^2}{3} =$$
$$\frac{a}{1 + 2b^3} + \frac{b}{1 + 2c^3} + \frac{c}{1 + 2d^3} + \frac{d}{1 + 2a^3} - a - b - c - d + \frac{2(ab + bc + cd + da + ac + bd)}{3} \ge 0$$
$$\frac{a}{1 + 2b^3} + \frac{b}{1 + 2c^3} + \frac{c}{1 + 2d^3} + \frac{d}{1 + 2a^3} - a - b - c - d + \frac{2(ab + bc + cd + da)}{3} =$$
$$a \left( \frac{1}{1 + 2b^3} - 1 + \frac{2b}{3} \right) + b \left( \frac{1}{1 + 2c^3} - 1 + \frac{2c}{3} \right) + c \left( \frac{1}{1 + 2d^3} - 1 + \frac{2d}{3} \right) +$$
$$d \left( \frac{1}{1 + 2a^3} - 1 + \frac{2a}{3} \right) =$$
$$\frac{2ab(2b^3 - 3b^2 + 1)}{3(1 + 2b^3)} + \frac{2bc(2c^3 - 3c^2 + 1)}{3(1 + 2c^3)} + \frac{2cd(2d^3 - 3d^2 + 1)}{3(1 + 2d^3)} + \frac{2ad(2a^3 - 3a^2 + 1)}{3(1 + 2a^3)} =$$
$$\frac{2ab(2b + 1)(b - 1)^2}{3(1 + 2b^3)} + \frac{2bc(2c + 1)(c - 1)^2}{3(1 + 2c^3)} + \frac{2cd(2d + 1)(d - 1)^2}{3(1 + 2d^3)} +$$
$$\frac{2ad(2a + 1)(a - 1)^2}{3(1 + 2a^3)} \ge 0.$$
We have equality when $ac + bd = 0$ and also $a = 3$, $b = c = d = 0$ or $a = 2$, $b = 1$, $c = d = 0$ as well as for their cyclic permutations.

Second solution. (Ionuț Nicolae) We use the inequality $\frac{1}{1 + 2a^3} \ge 1 - \frac{2a}{3}$ which, after computations, comes to $a(a - 1)^2(2a + 1) \ge 0$. The last inequality is obviously true and its equality cases are $a = 0$ and $a = 1$. We have $\frac{d}{1 + 2a^3} \ge \frac{d - 2ad}{3}$, with equality if $d = 0$ or $a \in \{0, 1\}$.
Then $\sum_{cycl} \frac{a}{1 + 2b^3} \ge \sum_{cycl} \left(a - \frac{2ab}{3}\right) = 3 - \frac{2}{3} \sum_{cycl} ab$. It remains to be proven that $a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + da) \le 9$, which follows from $a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + da + ac + bd) = 9$. We have equality if $ac + bd = 0$ and each non-zero variable is followed in cyclic order by one that is either $0$ or $1$. From $ac + bd = 0$ we see that two variables need to be $0$. If another one is $0$ then the last one is $3$ and we have indeed equality; if the other two are positive, the second one needs to be $1$, hence the first one must be $2$.

Third solution. (given in the contest by Andrei Pantea) From AM-GM inequality we have $1 + 2x^3 = 1 + x^3 + x^3 \ge 3x^2, \forall x > 0$.
We treat the following cases: 1. $a$, $b$, $c$, $d > 0$; 2. $a$, $b$, $c > 0$, $d = 0$; 3. $a$, $b > 0$, $c = d = 0$; 4. $a$, $c > 0$, $b = d = 0$; 5. $a = b = c = 0$, $d = 3$. All the other cases follow by cyclic permutations.
Case 1. The inequality is equivalent to
$$\sum_{cycl} \left( a - \frac{a}{1 + 2b^3} \right) \le \frac{1}{3} \left( (a + b + c + d)^2 - \sum_{cycl} a^2 \right).$$
But
$$\sum_{cycl} \frac{2ab^3}{1 + 2b^3} \le \sum_{cycl} \frac{2ab^3}{3b^2} = \frac{2}{3} \sum_{cycl} ab < \frac{2}{3} \sum_{cycl} ab = \frac{1}{3} \left( (a + b + c + d)^2 - \sum_{cycl} a^2 \right),$$
so in this case the inequality is satisfied without equality.