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Given a positive integer $n$, find the minimum value of $$\frac{x_1^3+x_2^3+...+x_n^3}{x_1+x_2+...+x_n}$$ subject to the condition that $x_1,x_2,...,x_n$ be distinct positive integers

I tried and this is what happened Order the numbers $x_1 < x_2 < ··· < x_n$ and call the expression from the statement $E(x_1, x_2,...,x_n)$. Note that $E(x1, x_2,...,x_n) > x^2_n/n$ , which shows that as the variables tend to infinity, so does the expression. This means that the minimum exists. Assume that the minimum is attained at the point $(y_1, y_2,...,y_n)$. If $y_n − y_1 > n$ then there exist indices $i$ and $j$ , $i<j$ , such that $y_1,...,y_{i}+1,...,y_j − 1,...,y_n$ are still distinct integers.

Dmitry
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2 Answers2

1

I think you have got the important first step down in your post. Say the minimum is attained at $x_1<x_2<\dots <x_n$ and assume that $x_n-x_1> n.$ Then like you said, there exists $1\leq i<j\leq n$ such that $x_1<x_2<\dots x_i+1<x_{i+1}<\dots x_{j-1}<x_j-1<\dots x_n.$ Call this new tuple $(y_1,y_2,\dots y_n)$ and observe that: $$E(y_1,y_2,\dots y_n) - E(x_1,x_2,\dots x_n) = \dfrac{(x_i+1)^3+(x_j-1)^3 - x_i^3-x_j^3}{\sum x_i} = \dfrac{3(x_i^2-x_j^2)+3(x_i+x_j)}{\sum x_i} = $$ $$ = \dfrac{3(x_i+x_j)(x_i-x_j+1)}{\sum x_i} < 0$$ since $x_j-1>x_i+1\implies x_i-x_j+1<-1.$

This severely restricts the possibilities into either $x_n = x_1+n$ or $x_n = x_1+n-1$, both of which should be easily bashable.

EDIT: First consider $x_n = x_1+n-1.$ This means, $x_k = a+(k-1), \,1\leq k\leq n.$ The denominator is easily $x_1+x_2+\dots+x_n = na+\frac{n(n-1)}{2}$ and the numerator is: $$\sum_{k=0}^{n-1}(a+k)^3 = na^3+\dfrac{3n(n-1)}{2}a^2+\dfrac{(n-1)n(2n-1)}{2}a+\dfrac{n^2(n-1)^2}{4}$$ and the whole expression miraculously turns out to be: $$E = \dfrac{na^3+\dfrac{3n(n-1)}{2}a^2+\dfrac{(n-1)n(2n-1)}{2}a+\dfrac{n^2(n-1)^2}{4}}{na+\frac{n(n-1)}{2}} = a^2+(n-1)a + \dfrac{n(n-1)}{2}$$ and so clearly, the minimum is attained at $a = 1,$ which would be $E = \dfrac{n(n+1)}{2}.$

Now, if $x_n = x_1 + n,$ then we can assume our tuple is the whole $a,a+1,\dots a+n$ except for one $a+j.$ The denominator is then $(n+1)a + \frac{n(n+1)}{2} - a - j = na+\dfrac{n(n+1)}{2}-j$ and the numerator is: $$\sum_{k=0}^n(a+k)^3 - (a+j)^3=na^3+3\left(\dfrac{n(n+1)}{2}-j\right)a^2+3\left(\dfrac{(n+1)n(2n+1)}{6}-j^2\right)a+\dfrac{n^2(n+1)^2}{4}-j^3$$ and the rest should follow similarly with a bit more algebra.

dezdichado
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0

Mathematical induction: $n=2$ and $\frac{1^3+2^3}{1+2}\lt \frac{1^3+k^3}{1+k}$ for $k\gt 2$ followed by $\frac{\sum\limits_{m=1}^{n+1}m^3}{\sum\limits_{m=1}^{n+1}m}\lt\frac{\sum\limits_{m=1}^nm^3+k^3}{\sum\limits_{m=1}^n m+k}$ for $k\gt n+1$.

herb steinberg
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