Let $a, b, c, d$ be real numbers such that $b-d \geq 5$ and all zeros $x_{1}, x_{2}, x_{3}$, and $x_{4}$ of the polynomial $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$ are real. Find the smallest value the product $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right)\left(x_{3}^{2}+1\right)\left(x_{4}^{2}+1\right)$ can take.
The answer is 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:
Claim - We always have $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right)\left(x_{3}^{2}+1\right)\left(x_{4}^{2}+1\right)=(b-d-1)^{2}+(a-c)^{2}$.

Proof. Let $i=\sqrt{-1}$. The key observation is that
$$
\prod_{j=1}^{4}\left(x_{j}^{2}+1\right)=\prod_{j=1}^{4}\left(x_{j}-i\right)\left(x_{j}+i\right)=P(i) P(-i)=|P(i)|^{2} .
$$

Since $P(i)=(-1+b-d)+(c-a) i$, the claim follows.
Since $b-d-1 \geq 4$, we get the desired lower bound of $4^{2}+0^{2}=16$.