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

## Comments

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TopNewestI 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 · 1 year, 9 months ago

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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 · 1 year, 9 months ago

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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 · 1 year, 9 months ago

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