# 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 $x_1x_2\dots x_n\le1$.

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.

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 $$|x_1 \cdots x_n| \leq 1$$.

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 $$\color{red}{x_2<0}$$.
$$({x_1}^2-x_1)+({x_2}^2-x_2)+({x_3}^2-x_3)=0$$
$$x_3-{x_3}^2=({x_1}^2-x_1)+({x_2}^2-x_2)>0$$
$$\therefore \color{red}{0
The three red inequalities imply $$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 $$1$$.
$$\therefore x_1x_2\dots x_n\le1$$

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: $$\left(\prod_{i=1}^n x_i \right)^{1/n}\leq 1.$$

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.