An inequality involving sums of squares and sums of cubes

Question

Assume that we have $m$ real numbers $x_1,x_2,...,x_m\in[0,1/4]$ satisfying the following equations: $$ \sum_{i=1}^m x_i=a_1, \sum_{i=1}^mx_i^2=a_2,\sum_{i=1}^mx_i^3\geq a_3,$$ where $a_1,a_2,a_3$ are all positive. The aim is to obtain restrictions on $a_1,a_2,a_3$. We can consider the polynomial $$f_b(x)=(x-1/4)(x+b)^2=x^3+(2b-1/4)x^2+(b^2-b/2)x-(1/4)b^2$$ where $b\in\mathbb{R}$. Then $$0\geq \sum_{i=1}^m f_b(x_i)\geq a_3+(2b-1/4)a_2+(b^2-b/2)a_1-(1/4)b^2m,$$ where the right hand side can be seen as a quadratic polynomial in variable $b$ and it is always less than or equal to 0. This will give us a restriction on $a_1,a_2,a_3$. I wonder whether there are other restrictions on $a_1,a_2,a_3$?

Answer 1

This is a partial answer. If it is correct, I can proceed.

Let $m \ge 3$ be a fixed positive integer. Let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$.

First, let us prove the following lemma.

Lemma 1: There exists $x_1, x_2, \cdots, x_m \in [0, \frac{1}{4}]$, not all zero, such that $x_1 + x_2 + \cdots + x_m = a_1, \ x_1^2 + x_2^2 + \cdots + x_m^2 = a_2$ and $x_1^3+x_2^3 + \cdots + x_m^3 \ge a_3$,

if and only if, \begin{equation} 0 < a_1 \le \frac{m}{4}, \qquad\qquad\qquad\qquad\qquad\qquad\qquad (1) \end{equation} \begin{equation} \frac{a_1^2}{m} \le a_2 \le \frac{1}{16}\lfloor 4a_1\rfloor + \left(a_1 - \frac{1}{4}\lfloor 4a_1\rfloor\right)^2, \qquad (2) \end{equation} and \begin{align} a_3 \le F(a_1, a_2),\qquad\qquad\qquad\qquad\qquad\qquad\quad (3) \end{align} where \begin{align} F(a_1, a_2) = &\max_{x_1, x_2, \cdots, x_m} \ x_1^3 + x_2^3 + \cdots + x_m^3\qquad\qquad\qquad\qquad\quad (4)\\ &\mathrm{s.t.}\quad x_i\in [0, \frac{1}{4}], \forall i; \ x_1 + x_2 + \cdots + x_m = a_1, \ x_1^2 + x_2^2 + \cdots + x_m^2 = a_2.\nonumber \end{align}

Proof of Lemma 1: WLOG, assume that $x_1 \ge x_2\ge \cdots \ge x_m$.

(the "only if" part):

i) Clearly, $0 < a_1 \le \frac{m}{4}$.

ii) Given $0 < a_1 \le \frac{m}{4}$, the range of $a_2$ is $[g(a_1), f(a_1)]$ where \begin{align*} f(a_1) = &\max_{x_1, x_2, \cdots, x_m} \ x_1^2 + x_2^2 + \cdots + x_m^2\\ &\mathrm{s.t.}\quad x_i\in [0, \frac{1}{4}], \forall i; \ x_1 + x_2 + \cdots + x_m = a_1 \end{align*} and \begin{align*} g(a_1) = &\min_{x_1, x_2, \cdots, x_m} \ x_1^2 + x_2^2 + \cdots + x_m^2\\ &\mathrm{s.t.}\quad x_i\in [0, \frac{1}{4}], \forall i; \ x_1 + x_2 + \cdots + x_m = a_1. \end{align*} It is easy to obtain $g(a_1) = \frac{a_1^2}{m}$ (the minimum is attained always when $x_1=x_2 = \cdots = x_m = \frac{a_1}{m}\in [0, \frac{1}{4}]$).

We have $f(a_1) = \frac{1}{16}\lfloor 4a_1\rfloor + \left(a_1 - \frac{1}{4}\lfloor 4a_1\rfloor\right)^2$ since $h(x) = x_1^2 + x_2^2 + \cdots + x_n^2$ is convex, and (denote $K = \lfloor 4a_1\rfloor$)

$\big(\underbrace{\frac{1}{4}, \frac{1}{4}, \cdots, \frac{1}{4}}_{K}, a_1 - \frac{1}{4}K, 0, 0, \cdots, 0\big)$ majorizes $(x_1, x_2, \cdots, x_m)$ with $x_i\in [0, \frac{1}{4}], \forall i; \ x_1 + x_2 + \cdots + x_m = a_1$.

iii) Clearly, (3) is true.

(the "if" part): It is clear according to the (the "only if" part). This completes the proof of Lemma 1.

Let us proceed. We want to find the explicit expression of $F(a_1, a_2)$.

Let $(x_1, \ x_2, \cdots, x_m)$ be the global maximizer of the optimization problem given in (4). We claim that $|\{x_1, \ x_2, \cdots, x_m\}|\le 4$ where $|\cdot|$ is the cardinality of a set.

Proof of the claim: Assume, for the sake of contradiction, that $|\{x_1, \ x_2, \cdots, x_m\}| > 4$. WLOG, assume that $0 < x_1 < x_2 < x_3 < \frac{1}{4}.$ Denote $D = 16(x_1^2+x_2^2+x_3^2) - 32(x_1x_2 + x_1x_3 + x_2x_3) + 8(x_1+x_2+x_3) -3.$

Remark: The proof of the inequalities in this part is simple and thus omitted.

1) If $D\le 0$, let $$a = \frac{x_1+x_2+x_3}{3}-\frac{1}{3}\sqrt{x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3}\, ,$$ $$b = \frac{x_1+x_2+x_3}{3}+\frac{2}{3}\sqrt{x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3}\, .$$ We have $x_1+x_2+x_3 = a + a + b, \ x_1^2+x_2^2+x_3^2 = a^2 + a^2 + b^2$ and $0 \le a \le b\le \frac{1}{4}$. Thus, $(a, a, b, x_4, x_5, \cdots, x_m)$ is a feasible point of the optimization problem given in (4). We also have $a^3 + a^3 + b^3 > (x_1^3 + x_2^3 + x_3^2) \Longleftarrow (x_2-x_3)^2(x_1-x_3)^2(x_1-x_2)^2 > 0.$ Thus, we have $a^3 + a^3 + b^3 + x_4^3 + x_5^3 + \cdots + x_m^3 > x_1^3 + x_2^3 + x_3^3 + x_4^3 + x_5^3 + \cdots + x_m^3.$ Thus, $(a, a, b, x_4, x_5, \cdots, x_m)$ is strictly better than $(x_1, \ x_2, \cdots, x_m)$. This contradicts the optimality of $(x_1, \ x_2, \cdots, x_m)$.

2) If $D > 0$, let $$u = \frac{x_1+x_2+x_3}{2} - \frac{1}{8} + \frac{1}{8}\sqrt{D}\,,$$ $$v = \frac{x_1+x_2+x_3}{2} - \frac{1}{8} - \frac{1}{8}\sqrt{D}\,.$$ We have $x_1 + x_2 + x_3 = \frac{1}{4} + u + v, \ x_1^2+x_2^2+x_3^2 = (\frac{1}{4})^2 + u^2 + v^2.$ It follows from $D>0$ and $0 < x_1 < x_2 < x_3 < \frac{1}{4}$ that $$x_1+x_2+x_3\ge \frac{1}{4},$$ $$x_1^2+x_2^2+x_3^2 \ge \frac{1}{16},$$ $$16(x_1x_2+x_1x_3+x_2x_3) - 4(x_1+x_2+x_3) + 1\ge 0,$$ $$16(x_1x_2 + x_1x_3 + x_2x_3) - 8(x_1+x_2+x_3) + 3 > 0,$$ $$-192x_1x_2x_3 + 48(x_1x_2+x_1x_3+x_2x_3) - 12(x_1+x_2+x_3) + 3 > 0.$$ With these facts, we have $0 \le v \le u \le \frac{1}{4}$ and $(\frac{1}{4})^3 + u^3 + v^3 > x_1^3 + x_2^3 + x_3^3.$ Thus, $(\frac{1}{4}, u, v, x_4, x_5, \cdots, x_m)$ is a feasible point of the optimization problem given in (4) and is strictly better than $(x_1, \ x_2, \cdots, x_m)$. This contradicts the optimality of $(x_1, \ x_2, \cdots, x_m)$. This completes the proof of the claim.