Let's determine the minimum possible value of the expression

$$
4+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}
$$

where $(x, y)$ runs through the set of all real number pairs.
Apply the inequality between the arithmetic mean and the geometric mean to the numbers $1, x^{2} y^{4}$, and $x^{4} y^{2}$:

$$
\frac{1+x^{2} y^{4}+x^{4} y^{2}}{3} \geq \sqrt[3]{1 \cdot x^{2} y^{4} x^{4} y^{2}}=x^{2} y^{2}
$$

which, when rearranged, means that

$$
4+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2} \geq 3
$$

It remains to be shown whether equality can hold, and if so, for what values of $x$ and $y$. In the inequality used, equality holds if $1=x^{2} y^{4}=x^{4} y^{2}$. This implies $x \neq 0, y \neq 0$, then $x^{2}=y^{2}$, and thus $x^{2}=y^{2}=1$, meaning $x= \pm 1, y= \pm 1$. In this case, it is immediately clear that equality indeed holds. This proves that the smallest possible value of the expression is 3, which is achieved when $x= \pm 1, y= \pm 1$.

Based on the work of Szabados Péter (Dombóvár, Illyés Gy. Gimn., II. o.t.)

Note. Another thought process, seen in many submissions, is as follows. Instead, let's find the minimum of $x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}$. Since, for example, $x=y=1$ makes the expression negative, the minimum must also be negative, so it suffices to find the maximum of

$$
3 x^{2} y^{2}-x^{4} y^{2}-x^{4} y^{2}=x^{2} y^{2}\left(3-x^{2}-y^{2}\right)
$$

among pairs $(x, y)$ such that $3-x^{2}-y^{2}>0$. If we apply the inequality between the arithmetic mean and the geometric mean to the triplet $x^{2}, y^{2}, 3-x^{2}-y^{2}$, we also reach our goal.