Question: For $n\ge 3$ determine all real solutions of the system of $n$ equations: $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
My approach: It is given that $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
Define $$S_n:=x_1+x_2+\cdots+x_n.$$
This implies that $$x_1+x_2+\cdots+x_{n-1}+x_n=x_n+\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}++x_i+x_{i+1}+\cdots+x_n=x_i+\frac{1}{x_i}\\ \cdots \\ x_1+x_2+\cdots+x_{n-1}+x_n=x_1+\frac{1}{x_1}.$$
Therefore, we have $$x_j+\frac{1}{x_j}=S_n, \forall 1\le j\le n.$$
Now for any $1\le j\le n,$ we have $$x_j+\frac{1}{x_j}=x_n+\frac{1}{x_n}\\ \implies \frac{1}{x_j}-\frac{1}{x_n}=x_n-x_j\\ \implies \frac{x_n-x_j}{x_nx_j}=x_n-x_j\\ \implies (x_n-x_j)\left(\frac{1}{x_nx_j}-1\right)=0\\ \implies x_j=x_n \text{ or } x_j=\frac{1}{x_n}.$$
Now let us have $i (0\le i\le n-1)$ of the $(n-1)$ numbers $x_j, 1\le j\le n-1$ such that $$x_j=x_n$$ and the rest $(n-1-i)$ numbers such that $$x_j=\frac{1}{x_n}.$$
Therefore, since $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\\implies i.x_n+(n-1-i).\frac{1}{x_n}=\frac{1}{x_n}\\ \implies i.x_n^2+(n-2-i)=0\\\implies i.x_n^2=2+i-n.$$
Now since $x_n\in\mathbb{R}$ and $x_n\neq 0\implies x_n^2>0.$ Now since $i\ge 0\implies i.x_n^2\ge 0 \implies 2+i-n\ge 0\implies i\ge n-2.$ Therefore $i=n-2,n-1$.
Now when $i=n-2$, we have $(n-2)x_n^2=0.$ Now since it is given that $n\ge 3\implies n-2\ge 1>0$. Therefore $x_n^2=0\implies x_n=0$. But $x_n\neq 0$. Therefore $i\neq n-2$.
Now when $i=n-1$, we have $(n-1)x_n^2=1\implies x_n^2=\frac{1}{n-1}\implies x_n=\pm\frac{1}{\sqrt{n-1}}.$ Therefore we have $$x_1=x_2=\cdots=x_n=\pm\frac{1}{\sqrt{n-1}}.$$
Therefore the required set of solutions are $$(x_1,x_2,\cdots,x_n)=\left(\frac{1}{\sqrt{n-1}},\frac{1}{\sqrt{n-1}},\cdots,\frac{1}{\sqrt{n-1}}\right),\left(-\frac{1}{\sqrt{n-1}},-\frac{1}{\sqrt{n-1}},\cdots,-\frac{1}{\sqrt{n-1}}\right).$$
Can someone check if my solution is correct or not? And if correct, is there a more better and efficient solution than this?
I get the same result. Here's my work.
I abhore "..."s, so I'll write the equations like this:
For $i=1$ to $n$,
$\sum_{k=1, k\ne i}^n x_k =\dfrac1{x_i} $.
Filling in the missing term,
$\sum_{k=1}^n x_k =\dfrac1{x_i}+x_i $.
Letting $S = \sum_{k=1}^n x_k$, we have $S = \dfrac1{x_i}+x_i $ so $x_i^2-Sx_i+1 = 0$ or $x_i = \dfrac{S\pm\sqrt{S^2-4}}{2} $.
Also, for $i \ne j$, $\dfrac1{x_i}+x_i = \dfrac1{x_j}+x_j $ or, multiplying by $x_ix_j$, $x_j-x_i^2x_j =x_i-x_ix_j^2 $ or $x_i-x_j =x_ix_j^2-x_i^2x_j =x_ix_j(x_j-x_i) $.
If $x_i \ne x_j$ then $x_ix_j = -1$ so $x_j = -\dfrac1{x_i} $.
Set $i = n$ so the other values are either $x_n$ or $-\dfrac1{x_n} $.
Suppose $m$ of them are $x_n$. Then $S =mx_n-(n-m)\dfrac1{x_n} $ so $x_n+\dfrac1{x_n} =mx_n-(n-m)\dfrac1{x_n} $ or, writing $x$ for $x_n$, $(m-1)x=(n-m+1)\dfrac1{x} $.
We can't have $m=1$ for then $\dfrac1{x} = 0$.
Therefore $m \ge 2$ so that $(m-1)x^2=(n-m+1) $ or $x =\pm\sqrt{\dfrac{n-m+1}{m-1}} $.
I will choose $x =\sqrt{\dfrac{n-m+1}{m-1}} $ for now.
Then
$\begin{array}\\ S &=m\sqrt{\dfrac{n-m+1}{m-1}}-(m-n)\sqrt{\dfrac{m-1}{n-m+1}}\\ &=x+\dfrac1{x}\\ &=\sqrt{\dfrac{n-m+1}{m-1}}+\sqrt{\dfrac{m-1}{n-m+1}}\\ \end{array} $
so, multiplying by $\sqrt{(n-m+1)(m-1)} $,
$m(n-m+1)-(m-n)(m-1) =(n-m+1)+(m-1) $ or $n =mn-m^2+m-(m^2-(n+1)m+n) =2m(n+1)-2m^2-n $ or $0 =2(m(n+1)-m^2-n) $ or $0 = m^2-(n+1)m+n $ or $0 =(m-1)(m-n) $.
Since $m > 1$, we have $m = n$ so all the $x_i$ are the same, so $S = \dfrac{n}{\sqrt{n-1}} $ and each $x_i =\dfrac1{\sqrt{n-1}} $.
Note that $\dfrac1{\sqrt{n-1}} +\sqrt{n-1} =\dfrac{1+n-1}{\sqrt{n-1}} =\dfrac{n}{\sqrt{n-1}} = S $.
If $x =-\sqrt{\dfrac{n-m+1}{m-1}} $ then all the signs are reversed, so the final result is the same.