2021 AMC 12A Problems/Problem 7
- The following problem is from both the 2021 AMC 10A #9 and 2021 AMC 12A #7, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Expand)
- 3 Solution 2 (Expand and then Factor)
- 4 Solution 3 (Beyond Overkill)
- 5 Video Solution (Simple & Quick)
- 6 Video Solution by Aaron He (Trivial Inequality)
- 7 Video Solution by North America Math Contest Go Go Go (Trivial Inequality, Simon's Favourite Packing Theorem)
- 8 Video Solution by Hawk Math
- 9 Video Solution by OmegaLearn (Trivial Inequality, Simon's Favorite Factoring)
- 10 Video Solution 6
- 11 Video Solution by TheBeautyofMath
- 12 Video Solution by The Learning Royal
- 13 Yet another Video Solution
- 14 See also
Problem
What is the least possible value of 
for real numbers 
and 
?
Solution 1 (Expand)
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
Solution 2 (Expand and then Factor)
We expand the original expression, then factor the result by grouping:
Clearly, both factors are positive. By the Trivial Inequality, we have ![\[\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=\boxed{\textbf{(D)} ~1}.\]](http://latex.artofproblemsolving.com/d/8/a/d8afd2da87874b8455fa5cdce076c030b2796580.png)
Note that the least possible value of occurs at 
~MRENTHUSIASM
Solution 3 (Beyond Overkill)
Like solution 1, expand and simplify the original equation to and let 
. To find local extrema, find where 
. First, find the first partial derivative with respect to x and y and find where they are 
:
![\[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\]](http://latex.artofproblemsolving.com/2/c/8/2c8883627988c35283b34d3dfa17825ed8bf491d.png)
![\[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\]](http://latex.artofproblemsolving.com/2/f/d/2fde4a923b8182ddc56a7bc835d6d6b8a438e351.png)
Thus, there is a local extremum at 
. Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning 
is the minimum of 
. Plugging into , we find 1 
.
~ DBlack2021
Video Solution (Simple & Quick)
~ Education, the Study of Everything
Video Solution by Aaron He (Trivial Inequality)
https://www.youtube.com/watch?v=xTGDKBthWsw&t=6m58s
Video Solution by North America Math Contest Go Go Go (Trivial Inequality, Simon's Favourite Packing Theorem)
https://www.youtube.com/watch?v=PbJK4KKfQjY&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=8
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn (Trivial Inequality, Simon's Favorite Factoring)
~ pi_is_3.14
Video Solution 6
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/s6E4E06XhPU?t=640 (for AMC 10A)
https://youtu.be/cckGBU2x1zg?t=95 (for AMC 12A)
~IceMatrix
Video Solution by The Learning Royal
Yet another Video Solution
See also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 10 Problems and Solutions | ||
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
