Let
$$g(x, y) = a + bx + cy + dxy - \frac{3}{1 + x^2 + y^2}.$$
Then taking the centre of the square $[-1, 1]^2$ and its four corners, we have
$$\begin{align*} g(0, 0) &= a - 3 \\ g(1, 1) &= a + b + c + d - 1 \\ g(-1, 1) &= a - b + c - d - 1 \\ g(-1, -1) &= a - b - c + d - 1 \\ g(1, -1) &= a + b - c - d - 1 \end{align*}$$

$$g(1, 1) + g(-1, 1) + g(-1, -1) + g(1, -1) = 4a - 4 = 4g(0, 0) + 8$$
We can consider two cases: either $g(0, 0) \le -1$ or $g(0, 0) > -1$.

In the first case, setting $(x, y) = (0, 0)$ gives the result claimed. Otherwise, we have $4g(0, 0) + 8 > 4$ in which case:
$$g(1, 1) + g(-1, 1) + g(-1, -1) + g(1, -1) > 4$$
Then (at least) one of the terms on the left hand side must exceed $1$ and the result is proved.