2021 AMC 12A Problems/Problem 7

Problem

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

$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

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 $\boxed{(D) 1}$ , which can be achieved at $x=y=0$ . ~aop2014

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

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2021 AMC 12A Problems/Problem 7

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Problem

Solution

Video Solution by Hawk Math

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