The claimed inequality is equivalent to $2ab + 4a + 4b + 8 \ge 2cd$, which can be written as
$$2ab + 4a + 4b + a^2 + b^2 + c^2 + d^2 + 4 \ge 2cd$$
on account of the condition $a^2 + b^2 + c^2 + d^2 = 4$. By the identity
$$a^2 + b^2 + 2ab + 4a + 4b + 4 = (a + b + 2)^2$$
we arrive at the equivalent and obvious inequality
$$(a+b+2)^2 + (c-d)^2 \ge 0$$
The case of equality occurs for
$$a + b = -2 \quad \text{and} \quad c = d$$
together with $a^2 + b^2 + c^2 + d^2 = 4$.
For instance when $a = b = c = d = -1$.