Well, I like 2048!

Algebra Level 4

Let a 1 , . . . , a 2048 a_1,...,a_{2048} be non-negative real numbers so that ∑ i = 1 2048 a i = 1 \sum _{ i=1 }^{ 2048 }{ { a }_{ i } }=1 . Find the maximum value of ∑ i = 1 2048 a i a i 2 + 1 \sum _{ i=1 }^{ 2048 }{ \frac { { a }_{ i } }{ { a }_{ i }^{ 2 }+1 } } .

If the answer can be written as x y \frac{x}{y} , where x , y x,y are positive integers so that ( x , y ) = 1 (x,y)=1 , find ∣ x − y ∣ \left| x-y \right| .

Bonus : Generalize!

Inspiration.


The answer is 1.

Steven Jim by Steven Jim , 17, Vietnam

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1 solution

Abhishek Sinha
Jun 30, 2018

Note that 0 ≤ a i ≤ 1 , ∀ i 0\leq a_i\leq 1, \forall i . The function f : [ 0 , 1 ] → R + f:[0,1]\to \mathbb{R}_+ defined by f ( x ) = x 1 + x 2 f(x)= \frac{x}{1+x^2} is concave. Hence, using Jensen's inequality , we have ∑ i = 1 2048 a i 1 + a i 2 ≤ 1 ( 1 / 2048 ) 2 + 1 = 204 8 2 1 + 204 8 2 , \sum_{i=1}^{2048} \frac{a_i}{1+a_i^2} \leq \frac{1}{(1/2048)^2+1}= \frac{2048^2}{1+2048^2}, where the equality holds when all a i a_i 's are equal. Thus the result follows.

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