2021 AMC 10A Problem 9

Expand both squares:
$$(xy-1)^2 + (x+y)^2 = x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 = x^2y^2 + x^2 + y^2 + 1.$$
The $xy$ terms cancel, and what remains factors:
$$x^2y^2 + x^2 + y^2 + 1 = (x^2 + 1)(y^2 + 1).$$
Each factor is at least $1$ because $x^2, y^2 \ge 0$, so the product is at least $1$. The value $1$ is attained at $x = y = 0$: $(0 - 1)^2 + 0^2 = 1$.

The answer is $\boxed{\textbf{(D)}\ 1}$.

A sum of two squares is $0$ only if both vanish simultaneously, which here is impossible ($xy = 1$ with $x = -y$ forces $x^2 = -1$). Expanding to look for cancellation is the standard move; the resulting $(x^2+1)(y^2+1)$ is the same identity behind $(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$.

Stop after expanding: $x^2y^2 + x^2 + y^2 + 1$ is a sum of three squares plus $1$, so it is at least $1$, with equality at $x = y = 0$. No factoring needed.

• Answering 0 by assuming both squares can vanish at once; xy = 1 and x = -y cannot hold together for real numbers.
• Trying to minimize by calculus or by setting $x = y$, which gives $(x^2-1)^2 + 4x^2 = x^4 + 2x^2 + 1 \ge 1$ but does not by itself prove the global minimum.