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In an answer to my question here, Yuri Negometyanov has some equations that I do not follow completely and he has not answered my comment-questions about them.

2.

At the same time, $A^2B^2 = (C-k)^2(C^2-(C-k)^2),$ with the least value at $k=1.$

Then $$A^4 B^4 = (C-1)^4(2C-1)^2 = (C-1)^4 (4C^2-4C+1),$$

$$P^4 = A^4B^4C^4 = \dfrac1{256}(4C^2-4C)^4(4C^2-4C+1)\\[4pt] > \dfrac1{256}\Big(4C^2-4C\Big)^5 = \dfrac1{256}\Big((2C-1)^2-1\Big)^5,$$ $$C < \dfrac12\left(\sqrt{(4P)^{^4/_5}+1\ }\ + 1\right).\tag2$$

Example: $A=35, B=12, C=37, P=15540, \dfrac12\left(\sqrt{62160^{0.8}+1}+1\right)\approx41.843.$

Where and how does $\frac{1}{256}$ enter the picture? And how does it turn into $\frac{1}{2}$ later?

poetasis
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1 Answers1

2

The ${1\over 256}$ and the ${1\over 2}$ are unrelated.

Re: your first question, a couple algebra steps are occurring at once.

  • First, we go from $$A^4B^4=(C-1)^4(4C^2-4C+1)$$ to $$A^4B^4C^4=C^4(C-1)^4(4C^2-4C+1),$$ just by multiplying both sides by $C^4$.

  • Now we work on the "$C^4(C-1)^4$" part of the right hand side. Specifically, we first have $$C^4(C-1)^4=(C(C-1))^4$$ by combining like powers, and then $$=(C^2-C)^4$$ by expanding.

  • Finally, we multiply by $4^4\over 4^4$: we have $$(C^2-C)^4={4^4\over 4^4}(C^2-C)^4={(4(C^2-C))^4\over 4^4}={(4C^2-4C)^4\over 256}$$ (where the second $=$ again involves combining like powers).

Re: your second question, note the "$2C$" on the right hand side of the inequality $$P^4>\dfrac1{256}\Big((2C-1)^2-1\Big)^5$$ (which is what we're left at just before getting the final inequality). In order to get an inequality with "$C$" on its own we're going to need to divide both sides of some equation by $2$ at some point.

Noah Schweber
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