Let us set
$$f(x, y) = (x + y)(x^2 + y^2).$$
We'll show that for any real $x$ the inequality $y \ge z$ implies $f(x, y) \ge f(x, z)$. After subtraction we see that
$$f(x, y) - f(x, z) = \frac{1}{2}(y - z)((x + y)^2 + (y + z)^2 + (z + x)^2) \ge 0.$$
Moreover, equality occurs when $y = z$ or $x = y = z = 0$, so either way it implies $y = z$.

We can rewrite the system (implicitly using the symmetry of $f$) to the form:
$$\begin{aligned} f(a, b) &= f(c, d) \\ f(a, c) &= f(b, d) \\ f(a, d) &= f(b, c) \end{aligned}$$
Now we can see that the system is symmetric in variables $a, b, c, d$ and may assume $a = \max\{a, b, c, d\}$. We then write the chain of (in)equalities
$$f(c, d) = f(a, b) \ge f(c, b) = f(a, d) \ge f(b, d) = f(a, c) \ge f(d, c)$$
and since we in fact have equality everywhere, we deduce $a = b = c = d$.
All such quadruplets clearly satisfy the system so the problem is solved. □