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The Inequality Inspired by Joel Tan: Generalization

See here and here for my spinoffs of Joel Tan's problem here.

Certainly, you may have realized that I didn't just randomly place exponents and get a new inequality every time; in fact, only certain triplets of exponents \((a, b, c)\) will make the maximum value of\[x (x+y)^{a}(y+z)^{b}(x+z)^{c}\] a rational number.

I created a computer program that will output all the ordered triplets that do make the maximum a rational number, along with the equality case \((x, y, z)\). Below is a list (in the range \(1\le a, b, c\le 10\) and with equality case omitted for obvious reasons). My challenge: make a computer program that does what my computer program does.

2, 2, 1

1, 2, 2

1, 3, 1

4, 3, 2

3, 3, 3

2, 3, 4\(\quad\leftarrow\text{ Joel's Inequality}\)

3, 4, 1

2, 4, 2

1, 4, 3

6, 4, 3

5, 4, 4\(\quad\leftarrow\text{ Spinoff 1}\)

4, 4, 5

3, 4, 6

4, 5, 2

3, 5, 3

2, 5, 4

8, 5, 4

7, 5, 5

6, 5, 6

5, 5, 7

4, 5, 8

2, 6, 1

4, 6, 1

1, 6, 2\(\quad\leftarrow\text{ Spinoff 2}\)

3, 6, 2

6, 6, 2

2, 6, 3

5, 6, 3

1, 6, 4

4, 6, 4

3, 6, 5

10, 6, 5

2, 6, 6

9, 6, 6

8, 6, 7

7, 6, 8

6, 6, 9

5, 6, 10

7, 7, 3

6, 7, 4

5, 7, 5

4, 7, 6

3, 7, 7

10, 7, 8

9, 7, 9

8, 7, 10

5, 8, 1

4, 8, 2

3, 8, 3

9, 8, 3

2, 8, 4

8, 8, 4

1, 8, 5

7, 8, 5

6, 8, 6

5, 8, 7

4, 8, 8

3, 8, 9

3, 9, 1

2, 9, 2

8, 9, 2

1, 9, 3

7, 9, 3

6, 9, 4

10, 9, 4

5, 9, 5

9, 9, 5

4, 9, 6

8, 9, 6

3, 9, 7

7, 9, 7

2, 9, 8

6, 9, 8

5, 9, 9

4, 9, 10

6, 10, 1

5, 10, 2

7, 10, 2

4, 10, 3

6, 10, 3

3, 10, 4

5, 10, 4

2, 10, 5

4, 10, 5

1, 10, 6

3, 10, 6

10, 10, 6

2, 10, 7

9, 10, 7

8, 10, 8

7, 10, 9

6, 10, 10

Note by Daniel Liu
2 years ago

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I don't quite like splitting out the AM-GM into 2 inequalities and multiplying them, as it is hard to justify the initial steps from a problem solver POV, instead of a problem creator POV. See my note on your other solution to see how all of these "magical constants" can be determined. Calvin Lin Staff · 2 years ago

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@Calvin Lin Actually, your method of solving for those magic constants is precisely the intended solution of solving both of my spinoffs (I had to do it myself too, after all) and it is also a key step in creating a reasonably fast computer program to generate all triples \((a,b,c)\) (at least it is my method).

As for my solution, I guess I just wanted people to be blown away on the arbitrariness of my division and how it just all perfectly works out. I guess I could have wrote a solution on solving for that division, but I didn't want to give everything away just yet. Daniel Liu · 2 years ago

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@Daniel Liu Right, I prefer for a solution to be instructive and illuminate a certain way of thinking/understanding. This allows us to get at the underlying ideas in a problem, and also shows how we can extend it to other problems even of different types.

E.g. What I outlined is a good approach to maximize the product of several linear terms, even if they are non-symmetric. The equations are easy to set up from the conditions in the problem, and then we're reduced to solving 2 systems of equations (separately) to find the values of the constants to use. Calvin Lin Staff · 2 years ago

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