WLOG $x>y$. the condition gives us $ x^{n-1}+x^{n-2}y+ \dots + y^{n-1}= x^{n}+x^{n-1}y+ \dots + y^{n}$ let $ x^{n-1}+x^{n-2}y+ \dots + xy^{n-2}=A$ and $y^{n-1}=B$ the condition become's $$A+B=Ax+(x+y)B$$and note that $$Ay+x^n+y^n=x^{n-1}+x^{n-2}y+ \dots + y^{n-1}=A+B$$. First partif $x+y \le 1$ from $A+B=Ax+(x+y)B$ we have $A+B \le Ax+B<A+B$ contradiction. therefore $x+y >1$. Second partAssume that $x+y=1+t$. we have $A+B=Ax+(1+t)B \implies A(y-t)=Bt$ then $Ay=(A+B)t$ now add $x^n+y^n$ to both sides, we get $$A+B= Ay+x^n+y^n=(A+B)t+x^n+y^n$$by $AM-GM$(which doesn't have equality case becuase $x>y$: $$(x^n+y^n)\frac{n+1}{2}> x^{n}+x^{n-1}y+ \dots + y^{n}=A+B$$So:$$ (x^n+y^n)> \frac{2}{n+1}(A+B)$$. And now we have: $$A+B=(A+B)t+x^n+y^n> (A+B)t+(A+B)\frac{2}{n+1}$$which implies :$$ A+B>(A+B)(t+\frac{2}{n+1}) \implies 1>t+\frac{2}{n+1}\implies \frac{n-1}{n+1} > t$$$$x+y=1+t < 1+ \frac{n-1}{n+1}=\frac{2n}{n+1} \implies x+y <\frac{2n}{n+1}$$And by First and Second part we have: $$1<x+y<\frac{2n}{n+1}$$