The answer is not − 1 -1

Algebra Level 5

Let a 1 , . . . , a 2018 a_1,...,a_{2018} be real numbers so that ∑ i = 1 2018 a i = 1 \sum _{ i=1 }^{ 2018 }{ { a }_{ i } }=1 . Find the maximum value of ∑ i = 1 2018 a i a i 2 + 1 \sum _{ i=1 }^{ 2018 }{ \frac { { a }_{ i } }{ { a }_{ i }^{ 2 }+1 } } .

If the answer can be written as x 2016 2 + 1 \frac{x}{{2016}^{2}+1} , find ⌊ 2 ∣ x − 2016 2 − 1 ∣ ⌋ \left \lfloor{2\left| x-{2016}^{2}-1 \right|}\right \rfloor .

Bonus : Generalize!

Inspiration.


The answer is 8189473823.

Steven Jim by Steven Jim , 17, Vietnam

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1 solution

X X
Jun 30, 2018

This isn't complete yet.

a 1 = − 2016 , a 2 = a 3 = . . . = a 2018 = 1 a_1=-2016,a_2=a_3=...=a_{2018}=1 ,so x = 8197602337 , y = 8128514 x=8197602337,y=8128514

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These values do not produce the maximum.

Set 2017 of the variables equal to x x . Then the last variable is 1 − 2017 x 1 - 2017x , and the sum is equal to f ( x ) = 2017 x x 2 + 1 + 1 − 2017 x ( 1 − 2017 x ) 2 + 1 . f(x) = \frac{2017x}{x^2 + 1} + \frac{1 - 2017x}{(1 - 2017x)^2 + 1}. Taking the derivative, we find f ′ ( x ) = − 2017 ( 2016 x − 1 ) ( 2018 x − 1 ) ( 4068289 x 4 − 4034 x 3 − 4068289 x 2 + 4034 x − 4 ) ( x 2 + 1 ) 2 ( 4068289 x 2 − 4034 x + 2 ) 2 . f'(x) = -\frac{2017(2016x - 1)(2018x - 1)(4068289x^4 - 4034x^3 - 4068289x^2 + 4034x - 4)}{(x^2 + 1)^2 (4068289x^2 - 4034x + 2)^2}. The polynomial 4068289 x 4 − 4034 x 3 − 4068289 x 2 + 4034 x − 4 4068289x^4 - 4034x^3 - 4068289x^2 + 4034x - 4 has a root that is very close to 1, but not equal to 1.

Jon Haussmann - 2 years, 11 months ago

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I see.I typed the equation in wolfram alpha,and it comes out to be x ≈ 1 x\approx 1 ,but I misread it as x = 1 x=1 .After I posted the solution,I found out,"Wait, x = 1 x=1 shouldn't be the solution!".But I couldn't think of why the answer is still 8189473823.Maybe @Steven Jim should edit it to integers instead of real numbers.

P.S.You are 1 of the solvers,too.How did you solve it?

X X - 2 years, 11 months ago

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Seen. Rechecking. Thanks.

Steven Jim - 2 years, 11 months ago

Relevant wiki: Jensen's Inequality

i think the answer is 201 8 2 201 8 2 + 1 \frac{2018^2}{2018^2+1}

consider the concave function f ( x ) = x x 2 + 1 x ∈ ( 0 , 1 ) using Jensen’s Inequality with x 1 + x 2 + ⋯ + x 2018 = 1 f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x 2018 ) 2018 ≤ f ( x 1 + x 2 + ⋯ + x 2018 2018 ) x 1 x 1 2 + 1 + x 2 x 2 2 + 1 + ⋯ + x 2018 x 2018 2 + 1 ≤ 2018 f ( 1 2018 ) x 1 x 1 2 + 1 + x 2 x 2 2 + 1 + ⋯ + x 2018 x 2018 2 + 1 ≤ 201 8 2 201 8 2 + 1 \begin{aligned} &\text{consider the concave function } f(x)=\frac{x}{x^2+1} x \in (0,1)\\ &\text{using Jensen's Inequality with } x_1+x_2+\cdots +x_{2018}=1 \\ &\frac{f(x_1)+f(x_2)+\cdots +f(x_{2018})}{2018}\leq f\left ( \frac{x_1+x_2+\cdots +x_{2018}}{2018} \right )\\ &\frac{x_1}{x^2_1+1}+\frac{x_2}{x^2_2+1}+\cdots +\frac{x_{2018}}{x^2_{2018}+1}\leq 2018f\left ( \frac{1}{2018} \right )\\ &\frac{x_1}{x^2_1+1}+\frac{x_2}{x^2_2+1}+\cdots +\frac{x_{2018}}{x^2_{2018}+1}\leq \frac{2018^2}{2018^2+1} \end{aligned}

Hassan Abdulla - 2 years, 11 months ago

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It's not a concave function,since the variables can be negative.

X X - 2 years, 11 months ago

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