2021 AMC 10A Problem 9

This asks for a smallest value, so there are two halves: find values of $x$ and $y$ that reach it, then show nothing goes lower.

1. Try some easy numbers first:

So $1$ can be reached. Can anything go lower?
2. To find out, multiply out each square:
$$(xy - 1)^2 = x^2y^2 - 2xy + 1, \qquad (x + y)^2 = x^2 + 2xy + y^2.$$
3. Add them. The $-2xy$ and $+2xy$ cancel:
$$x^2y^2 + x^2 + y^2 + 1.$$
4. Each of $x^2y^2$, $x^2$ and $y^2$ is a square, so none of them can be negative. The whole thing is therefore at least $0 + 0 + 0 + 1 = 1$.
5. So $1$ can be reached (step 1) and cannot be beaten (step 4).

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

A sum of two squares can only be $0$ if both squares are $0$ at once. Here that would need $xy = 1$ and $x + y = 0$ together, which forces $x \times (-x) = 1$, and no real number does that. Multiplying out to look for cancelling terms is the standard first move with sums of squares.

The expression from step 3 also factors neatly:
$$x^2y^2 + x^2 + y^2 + 1 = x^2(y^2 + 1) + (y^2 + 1) = (x^2 + 1)(y^2 + 1).$$
Each factor is at least $1$, so the product is at least $1$, with $1$ reached when $x = y = 0$.

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 $\boxed{\textbf{(D)}\ 1}$.