2021 AMC 12A Spring 第 7 题

Question

对实数

x

和

y

，

(xy - 1)^2 + (x + y)^2

的最小可能值是多少？

What is the least possible value of

(xy - 1)^2 + (x + y)^2

for real numbers

x

and

y?

0

\dfrac14

\dfrac12

1

2

完全展开；交叉项

-2xy

和

+2xy

会抵消

Expand fully; the cross terms

-2xy

and

+2xy

cancel

表达式等于

(x^2 + 1)(y^2 + 1)

，两个因子都至少为

1

。

The expression equals

(x^2 + 1)(y^2 + 1),

a product of two factors each at least

1

展开得

\begin{aligned} &(xy-1)^2 + (x+y)^2 \\ &= x^2y^2 - 2xy + 1 + x^2 \\ &\quad {}+ 2xy + y^2 \\ &= x^2y^2 + x^2 + y^2 + 1\text{。} \end{aligned}

这可因式分解为

(x^2 + 1)(y^2 + 1)

。

每个因子都至少为

1

，所以乘积至少为

1

，且当

x = y = 0

时取到等号。

因此，正确答案是 D。

Expanding,

\begin{aligned} &(xy-1)^2 + (x+y)^2 \\ &= x^2y^2 - 2xy + 1 + x^2 \\ &\quad {}+ 2xy + y^2 \\ &= x^2y^2 + x^2 + y^2 + 1. \end{aligned}

This factors as

(x^2 + 1)(y^2 + 1).

Each factor is at least

1,

so the product is at least

1,

with equality when

x = y = 0.

Thus, the correct answer is D.

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