To analyze the inequality \( (a^2 + 1)(b^2 + 1) \geq 4ab \), let's simplify and prove it step by step.

We want to show that:

\[
(a^2 + 1)(b^2 + 1) \geq 4ab
\]

First, expand the left-hand side:

\[
a^2b^2 + a^2 + b^2 + 1 \geq 4ab
\]

Rearranging gives:

\[
a^2b^2 + a^2 + b^2 + 1 - 4ab \geq 0
\]

Next, this can be thought of as a quadratic in \( ab \). Let's denote \( x = ab \), then \( a^2b^2 = x^2 \). The inequality can be rewritten as:

\[
x^2 - 4x + (a^2 + b^2 + 1) \geq 0
\]

Now using the properties of quadratic equations, we will analyze the discriminant. The discriminant \( D \) must be non-positive for the quadratic to be non-negative (which means the roots are either real and distinct or complex, or it is a perfect square). We compute the discriminant:

\[
D = (-4)^2 - 4 \cdot 1 \cdot (a^2 + b^2 + 1)
\]
\[
= 16 - 4(a^2 + b^2 + 1)
\]
\[
= 16 - 4a^2 - 4b^2 - 4
\]
\[
= 12 - 4a^2 - 4b^2
\]

We want:

\[
D \leq 0
\]

Thus,

\[
12 - 4(a^2 + b^2) \leq 0
\]

which simplifies to:

\[
4(a^2 + b^2) \geq 12
\]
\[
a^2 + b^2 \geq 3
\]

By applying the AM-GM inequality, we know:

\[
\frac{a^2 + b^2}{2} \geq \sqrt{a^2b^2} = ab
\]

Also, by Cauchy-Schwarz inequality:

\[
(a^2 + b^2)(1 + 1) \geq (a + b)^2 \implies a^2 + b^2 \geq \frac{(a + b)^2}{2}
\]

In conclusion, it's required that \( a^2 + b^2 \geq 3 \) to satisfy the inequality \( (a^2 + 1)(b^2 + 1) \geq 4ab \). If \( a \) and \( b \) are both positive values, the provided conditions especially recognize the established relations from the quadratic and inequality proofs.

Thus, we have shown that:

\[
(a^2 + 1)(b^2 + 1) \geq 4ab
\]

is valid.