Difference between revisions of "2021 AMC 12A Problems/Problem 7"
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Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the trivial inequality(all squares are nonnegative) the minimum value for this is <math>\boxed{(D) 1}</math>, which can be achieved at <math>x=y=0</math>. ~aop2014 | Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the trivial inequality(all squares are nonnegative) the minimum value for this is <math>\boxed{(D) 1}</math>, which can be achieved at <math>x=y=0</math>. ~aop2014 | ||
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==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=6|num-a=8}} | {{AMC12 box|year=2021|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:56, 11 February 2021
Problem
What is the least possible value of 
for real numbers 
and 
?
Solution
Expanding, we get that the expression is 
or 
. By the trivial inequality(all squares are nonnegative) the minimum value for this is 
, which can be achieved at 
. ~aop2014
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
