Real cubes

Algebra Level 5

The sum of 2016 real numbers is 0. If none of the numbers exceeds 1, then what is the maximum possible sum of their cubes?

The answer is 504.

2 solutions

Miles Koumouris
Apr 14, 2017

Observe that the inequality $-4x^{3}+3x+1=(1-x)(2x+1)^{2}\geq 0$ holds $\forall$ $x\leq 1$ . Assigning the numbers as $x_{k}$ for some $1\leq k\leq 2016$ , it is evident that

$\displaystyle \sum_{k=1}^{2016}(-4x_{k}^3+3x_{k}+1)=2016-4\displaystyle \sum_{k=1}^{2016}x_{k}^3\geq 0$

$\Longrightarrow \displaystyle \sum_{k=1}^{2016}x_{k}^3\leq 504$ .

Such a sum is obtained if $x_{k}=1$ for $1\leq k\leq 672$ , and $x_{k}=\dfrac{-1}{2}$ for $673\leq k\leq 2016$ .

Hence, the answer is $\boxed{504}$ .

Kushal Bose
Apr 15, 2017

Let $\displaystyle \sum_{i=0}^{2016} x_i =0$ and each $x_i \leq 1$ .So, there will some positive values and some negative values.

While adding the cubes of the terms then there will be still negative values and positive values.

Assume there are $k$ positive terms and $2016-k$ negative terms.To maximize positive values we can assign $k$ terms will equal to $1$ .So, total value $k \times 1=k$ .So, rest of the values will be negative and should be $-k$ .So, minimum $x_i$ for $k+1 \leq i \leq 2016$ is $(2016-k) x_i=-k \implies x_i=\dfrac{-k}{2016-k}$

Now, $\displaystyle \sum_{i=0}^{2016} x^3_i =\displaystyle \sum_{i=0}^{k} x^3_i + \displaystyle \sum_{i=k+1}^{2016} x^3_i \\ =k+(2016-k) x^3_i \\ =k- \dfrac{k^3}{(2016-k)^2}$

Now, using simple calculus maximum value of $k=672$

Using this maximum value of sum is $504$

This is an incomplete solution, because it might be possible to achieve higher value by not setting all positive values as 1.

Ivan Koswara - 4 years, 1 month ago

if you set values strictly less than 1 then it may be fractions or negative numbers.SO, their cubes will be strictly negative and total positive value will be reduced.

Kushal Bose - 4 years, 1 month ago

But if you make some positive value less than 1, then the sum of the positive values becomes smaller. So the negative values also become smaller (in magnitude); this might potentially make the overall sum of cubes to be larger.

Ivan Koswara - 4 years, 1 month ago

The largest fraction that can cancel 1 is 1/2. Two of "-1/2" will cancel one "1" , and in three terms, when cubed, we get 1-2/8=3/4. \therefore the average per 2016 terms =(3/4)/3=1/4.
For 2016 terms, the sum = 2016 * 1/4=504.

$If~ we~ take~ a ~fraction~~(i)~~greater~ than~ or~~(ii)~~ less~ than ~\dfrac 1 2,\\ say ~\dfrac 1 2~ \pm ~\dfrac 1 n=\dfrac {n \pm 2}{2n}.\\ ~ In~ order~ to ~cancel~~ "1s", ~~this~ fraction~\dfrac {n \pm 2}{2n} ~must~ have~ 2n ~terms.\\ So ~maximum ~~"1s"~~ must~ be ~(n ~\pm ~2) ~terms.\\ So~ in~( 3n~\pm ~2)~ terms,~ we ~will~ have ~(n\pm 2)~~ "1s"~~ and~ 2n~ of~~ "-\dfrac {n \pm 2}{2n}"~~ terms. \\ Sum~ of 3n\pm ~2~ terms ~=n\pm 2-(n\pm 2)=0.\\ Sum ~of~ the~ cubes~ for~ 3n\pm 2~ terms =n\pm 2- 2n * (\dfrac{n \pm 2}{2n})^3\\ =(n\pm 2)(1-(\dfrac{n\pm 2}{2n})^2=(n\pm 2)(1 -(\frac 1 2 \pm \frac 1 n )^2)=(n+2)(3/4 - \dfrac 1{n^2} \mp \frac 1 n ).\\ \therefore~ average~=\dfrac{n \pm 2}{3n\pm 2}*(3/4-\dfrac 1{n^2} \pm \frac 1 n)<1/4~ average ~ for~ 1/2. \\ Say~ if~ n~=~ \pm~ 4 ,~~average ~for~both,~\pm~is~ 3/16 ~from~above~formula.$

Niranjan Khanderia - 4 years, 1 month ago

Real cubes

The answer is 504.

2 solutions

0 pending reports