Inspired by Julian Poon and an encyclopedia

Algebra Level 5

Let x , y , z x,y,z be positive real numbers such that x 2 + y 2 + z 2 = 2 ( x y + y z + z x ) x^2 + y^2 + z^2 = 2(xy + yz+zx) . Find the maximum value of

A = x 26235 + y 26235 + z 26235 r = 1 53 ( x r ( r + 1 ) / 2 + y r ( r + 1 ) / 2 + z r ( r + 1 ) / 2 ) . A = \dfrac{ x^{26235} + y^{26235} + z^{26235}}{\displaystyle \prod_{r=1}^{53} \left ( x^{r(r+1)/2} + y^{r(r+1)/2} + z^{r(r+1)/2} \right) }.

Write your answer in the form of 10 6 A \left\lfloor { 10 }^{ 6 }A \right\rfloor .


Inspiration .


The answer is 646147.

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1 solution

Steven Jim
Jun 21, 2017

[Not a solution, as I'm stuck with Julian's problem]

The maximum value is 2 + 4 26235 ( 2 + 4 ) ( 2 + 4 3 ) . . . ( 2 + 4 1431 ) \frac { 2+{ 4 }^{ 26235 } }{ (2+4)(2+{ 4 }^{ 3 })...(2+{ 4 }^{ 1431 }) } .

Hey, just since I was meditating on the inspiration problem too, did you mean to write x 512656 + y 512656 + z 512656 {x}^{512656}+{y}^{512656}+z^{512656} as the numerator? (where 512656 = 71 6 2 = 1 + 3 + 5 + . . . + 1431 512656=716^2=1+3+5+...+1431 )

Mat Met - 3 years, 11 months ago

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Ah... No. What I wanted to mention is about the triangular numbers. I'll edit the problem. Thanks.

Steven Jim - 3 years, 11 months ago

Sorry for the mistake.

By the way, I've edited the page. Check out.

Steven Jim - 3 years, 11 months ago

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Thanks! Now it's clearer

Mat Met - 3 years, 11 months ago

Thank god @Julian Poon , at least someone got it correct XD

Steven Jim - 3 years, 11 months ago

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Ya, @Julian Poon reveal a very good solution, or a very good approximation to solve this kind of problem!

Kelvin Hong - 3 years, 11 months ago

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Now we need a full solution XD

Steven Jim - 3 years, 11 months ago

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@Steven Jim Ya, we need to approach something accurate XD

Kelvin Hong - 3 years, 11 months ago

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