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Show that if $\sum\limits_{k=1}^nx_k=\sum\limits_{k=1}^n{x_k}^2=\dots=\sum\limits_{k=1}^n{x_k}^n$ where $x_k\in\mathbb{R}$ then $\prod\limits_{k=1}^n x_k\le1$.

For example, when $n=3$, the system of equations is:

$$x_1+x_2+x_3$$ $$={x_1}^2+{x_2}^2+{x_3}^2$$ $$={x_1}^3+{x_2}^3+{x_3}^3$$

where $x_1, x_2, x_3 \in\mathbb{R}$. We are to show that $x_1 x_2 x_3 \le1.$

I made up this question. I have found a solution, which I will post, but my solution is rather complicated, and I am hoping for a more elegant solution.

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4 Answers 4

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Here is another solution, which only uses the following assumption:

Assumption. $n \geq 2$ and $x_1, \ldots, x_n \in \mathbb{R}$ satisfy $$ \sum_{k=1}^{n} x_k = \sum_{k=1}^{n} x_k^2. \tag{*} $$

Let $p$ denote the common value of the sums in $\text{(*)}$. Clearly, $p \geq 0$. Moreover, by Cauchy–Schwarz inequality,

\begin{align*} p = \sum_{k=1}^{n} x_k^2 = \frac{1}{n} \left( \sum_{k=1}^{n} x_k^2 \right)\left( \sum_{k=1}^{n} 1 \right) \geq \frac{1}{n} \left( \sum_{k=1}^{n} x_k \right)^2 = \frac{p^2}{n}. \end{align*}

Hence we get $0 \leq p \leq n$. Then by AM–GM inequality,

$$ |x_1 \cdots x_n|^{\frac{2}{n}} \leq \frac{x_1^2 + \cdots + x_n^2}{n} \leq 1, $$

which is enough to conclude that $|x_1 \cdots x_n| \leq 1$.

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We consider the cases $n=2, n=3$ and $n\ge4$.

$\color{blue}{n=2:}$
Let $p=x_1+x_2={x_1}^2+{x_2}^2$
${x_1}^2+(p-x_1)^2=p$
$2{x_1}^2-2px_1+(p^2-p)=0$
$\Delta=4p^2-8(p^2-p)\ge0$
$0\le p\le2$
$\therefore x_1x_2=\text{product of roots}=\frac{p^2-p}{2}\le1$

$\color{blue}{n=3:}$
Suppose $x_1x_2x_3>1$.
One of $|x_1|,|x_2|,|x_3|$ must be greater than $1$. Without loss of generality, assume $|x_1|>1$.
If $x_1<-1$ then $0=({x_1}^2-x_1)+({x_2}^2-x_2)+({x_3}^2-x_3)>2-1/4-1/4>0$, contradiction. $\therefore \color{red}{x_1>1}$
$({x_1}^3-2{x_1}^2+x_1)+({x_2}^3-2{x_2}^2+x_2)+({x_3}^3-2{x_3}^2+x_3)=0$
$x_1(x_1-1)^2+x_2(x_2-1)^2+x_3(x_3-1)^2=0$
So at least one of $x_2$ and $x_3$ must be negative. Without loss of generality, assume $\color{red}{x_2<0}$.
$({x_1}^2-x_1)+({x_2}^2-x_2)+({x_3}^2-x_3)=0$
$x_3-{x_3}^2=({x_1}^2-x_1)+({x_2}^2-x_2)>0$
$\therefore \color{red}{0<x_3<1}$
The three red inequalities imply $x_1x_2x_3<0$, contradiction.
$\therefore x_1x_2x_2\le 1$.

$\color{blue}{n\ge 4:}$
$({x_1}^4-2{x_1}^3+{x_1}^2)+({x_2}^4-2{x_2}^3+{x_2}^2)+\dots+({x_n}^4-2{x_n}^3+{x_n}^2)=0$
$(x_1(x_1-1))^2+(x_2(x_2-1))^2+\dots+(x_n(x_n-1))^2=0$
So every $x_k$ is either $0$ or $1$.
$\therefore x_1x_2\dots x_n\le1$

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Assume that the numbers are all positive, because if the number of negative numbers is even, then we can just replace a pair of negatives by a nonnegative, otherwise the product is negative or zero and the assertion follows immediately.

From the power-means inequalities, we have that for $1\leq k \leq n$: $$ \frac{1}{n}\sum_{i=1}^nx_i\leq\left(\frac{1}{n}\sum_{i=1}^n x_i^k\right)^{1/k}=\left(\frac{1}{n}\sum_{i=1}^n x_i\right)^{1/k} $$ or $$ \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^{1-1/k}\leq 1 \implies \frac{1}{n}\sum_{i=1}^n x_i \leq 1 $$ which by AM-GM: $$ \left(\prod_{i=1}^n x_i \right)^{1/n}\leq 1. $$

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Assuming the numbers are non-negative, this is pretty short and simple.

Case 1: $\sum_{k = 1}^n x_k \leq n$. Then by the AM-GM inequality, we have $$ \sqrt[n]{x_1x_2\cdots x_n}\leq \frac{x_1 + x_2 + \cdots + x_n}n \leq 1 $$ Raising everything to the $n$-th power yields the desired result.

Case 2: $\sum_{k = 1}^nx_k \geq n$. Then by the AM-QM inequality we have $$ \frac{x_1 + x_2 + \cdots + x_n}{n}\leq \sqrt{\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}} $$ and since the left-hand side and therefore the right-hand side is greater than or equal to $1$ we also have $$ \sqrt{\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}}\leq \frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n} $$ But this puts $\sqrt{\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}}$ in-between two terms we know are equal, which is to say, we must have $$ \sqrt{\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}} = \frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n} $$ This tells us that $$\frac{x_1+x_2+\cdots+x_n}n = \frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}=1$$Which takes us back to case 1.

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