$\sum x_{k}=1$ then, what is the maximal value of $\sum x_{k}^{2}\sum kx_{k} $

Let $1\geq x_{1}\geq x_2\geq\cdots\geq x_{n}\geq0$, and $\sum\limits_{k=1}^{n}x_{k}=1$. then what is the maximal value of ? $$\sum_{k=1}^{n}x_{k}^{2}\sum_{k=1}^{n}kx_{k} .$$

I think, Maybe we could try Rearrangement inequality

or

let $x_n=a_n,x_{n-1}=a_n+a_{n-1} , \dotsc, x_1=a_1+a_2+\dotsc+a_n$

where $a_i\geq0$

I am really grateful for any help

EDIT: I am very sorry, $1$ is not the upper bound for all $n$, but I guess $\sum\limits_{k=1}^{n}x_{k}^{2}\sum\limits_{k=1}^{n}kx_{k}\lt2 .$

For $n=5$ $$x_1=0.926599, x_2=x_3=x_4=x_5=0.0183503$$ LHS $=1.01773\gt 1$

Solution 1:

This function doesn't have a single maximum independent of $n$, given the constraints. Pick

$$X = \left(1-t,\frac{t}{n-1},\ldots,\frac{t}{n-1}\right)$$

with $n$ elements and $0 \leq t \leq 1/2$. Clearly $X$ satisfies your constraints. Now

$$g(X) = 2(n-1)f(X) = ((n-1)(1-t)^2+t^2)(2(1-t)+(n+2)t)$$

and both $g$ and $f$ attain a global maximum for the same $t$, given the constraints. Taking the first derivative of $g(X)$ w.r.t. $t$ and setting it to 0 yields the equation

$$2n^2t^2-4n(n-2)t+(n-1)(n-4) = 0$$

with solutions

$$t_{\pm} = \frac{2n-4\pm \sqrt{n^2-n+4}}{3n}$$

$t_{+}$ becomes too large at $n > 2$ and $t_{-1}$ becomes positive when $n>4$. Furthermore $t_{-1} < 1/2$ for all $n>4$ and

$$\lim_{t\to \infty} t_{-1} = \frac{1}{3}$$

So it is a valid substitution for $t$ for $n > 4$. After some simplification we obtain

$$f_{max} \ge \frac{(n+1)(n^2+11n-8)+(n^2-n+4)^{3/2}}{27n(n-1)}$$

and this quantity goes to infinity as n goes to infinity. Therefore, a simple bound like $f_{max} \leq 1$ or $f_{max} \leq 2$ will not suffice. In fact, the quantity so far is asymptotic to

$$\frac{(n+1)(n^2+11n-8)+(n^2-n+4)^{3/2}}{27n(n-1)} \sim \frac{2n}{27}$$

so for now you expect a bound of $f_{max} \ge cn$ for some constant $c$.

EDIT: Note that

$$1 = \sum_k x_k \ge n x_n$$

so $x_n \leq \frac{1}{n}$. In the case I described $x_n \sim \tfrac{1}{3n}$ as $n\to \infty$. So there is somewhat of a gap left between the case I described and the possible maximum.

EDIT 2: I believe that the maximum is given by the expression for $f_{max}$ above.

Solution 2:

This is not a complete answer, but is slightly too long for the usual comment format. Below is a proof for $n=3$ (which implies the result for any $n\leq 3$).

Following the suggestion in Winther’s comment, let us that $\delta\geq 0$, where

$$ \delta=(x_1+x_2+x_3)^3-(x_1^2+x_2^2+x_3^2)(x_1+2x_2+3x_3) \tag{1} $$

If we put $d_i=x_i-x_{i+1} (i=1,2)$, we have $d_i\geq 0$. The result then follows from

$$ \delta=x_2(x_1^2+x_2(x_1+d_1))+x_3(6x_1x_2+x_3(x_1+d_1+2d_2)) \tag{2} $$

Note that equality is reached exactly when $x_2=x_3=0$.