求代数式$x^2+y^2-x-y-xy$的最小值及可解性判断

Let's break this down step by step to find the minimum value and confirm if the problem is solvable.

Method 1: Completing the Square (Algebraic Approach) #

This is the most straightforward way to tackle this two-variable quadratic expression. Start by rearranging terms to set up completing the square:

$$
x^2 + y^2 - x - y - xy
$$

First, treat the expression as a quadratic in $x$:
$$
x^2 - (y + 1)x + (y^2 - y)
$$

Complete the square for the $x$-terms:
$$
\left(x - \frac{y+1}{2}\right)^2 - \left(\frac{y+1}{2}\right)^2 + y^2 - y
$$

Simplify the constant terms (those without $x$):
$$
-\frac{(y+1)^2}{4} + y^2 - y = \frac{-y^2 - 2y - 1 + 4y^2 - 4y}{4} = \frac{3y^2 - 6y - 1}{4}
$$

Now complete the square for the $y$-terms in this simplified part:
$$
\frac{3y^2 - 6y - 1}{4} = \frac{3(y^2 - 2y) - 1}{4} = \frac{3\left[(y-1)^2 - 1\right] - 1}{4} = \frac{3(y-1)^2}{4} - 1
$$

Putting it all together, the original expression rewrites to:
$$
\left(x - \frac{y+1}{2}\right)^2 + \frac{3(y-1)^2}{4} - 1
$$

Since squared terms are always non-negative ($\geq 0$), the minimum value occurs when both squared terms equal 0:

1. $\frac{3(y-1)^2}{4} = 0 \implies y = 1$
2. Substitute $y=1$ into $\left(x - \frac{y+1}{2}\right)^2 = 0$: $x - \frac{1+1}{2} = 0 \implies x = 1$

Plugging $x=1$ and $y=1$ back into the original expression:
$$
1^2 + 1^2 - 1 - 1 - (1)(1) = 1 + 1 - 1 - 1 - 1 = -1
$$

Is the Problem Solvable? #

Absolutely. We've found concrete values ($x=1, y=1$) that yield the minimum value of -1, and the non-negativity of squared terms confirms this is the smallest possible value the expression can take. There are no contradictions or undefined conditions here—this is a well-defined quadratic expression that attains a finite minimum.

内容的提问来源于stack exchange，提问作者Patrick Edward Corvera