Problem: Let $a$, $b$, $c$, $d$ be real numbers such that $b-d \ge 5$ and all zeros
$x_1$, $x_2$, $x_3$, and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real.
Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.

Solution: The answer is $\boxed{16}$.
This can be achieved by taking $x_1 = x_2 = x_3 = x_4 = 1$,
whence the product is $2^4 = 16$, and $b-d = 5$.

We now show the quantity is always at least $16$.
We prove:
\begin{claim*}
  We always have $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1) = (b-d-1)^2 + (a-c)^2$.
\end{claim*}
\begin{proof}
  Let $i = \sqrt{-1}$.
  The key observation is that
  \[ \prod_{j=1}^4 \left( x_j^2 + 1 \right)
    = \prod_{j=1}^4 (x_j - i)(x_j + i)
    = P(i)P(-i) = |P(i)|^2. \]
  Since $P(i) = (-1+b-d) + (c-a)i$, the claim follows.
\end{proof}
Since $b-d-1 \ge 4$, we get the desired lower bound of $4^2+0^2=16$.