<s>[INST] For $n\ge 3$ determine all real solutions of the system of $n$ equations. 
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?
 [/INST] 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.
</s>