AMC12 2021 A

AMC12 2021 A · Q7

AMC12 2021 A · Q7. It mainly tests Inequalities (AM-GM etc. basic).

What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$?

对于实数 $x$ 和 $y$，$(xy-1)^2+(x+y)^2$ 的最小可能值是多少？

(A) 0 0

(B) \frac{1}{4} \frac{1}{4}

(D) 1 1

(E) 2 2

Answer

Correct choice: (D)

正确答案：（D）

Solution

Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$. By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{\textbf{(D)} ~1}$, which can be achieved at $x=y=0$.

展开表达式，得 $x^2+2xy+y^2+x^2y^2-2xy+1$，即 $x^2+y^2+x^2y^2+1$。由平凡不等式（所有平方项非负），该表达式的值的最小值为 $\boxed{\textbf{(D)} ~1}$，可在 $x=y=0$ 时取得。

Topics

AL.013 Inequalities (AM-GM etc. basic)