$LHS-RHS=0$ $(x-1)\left[nx-\left(\frac{2}{x+1}+\frac{3}{x+2}+\ldots+\frac{n+1}{x+n}\right)\right]=0$ So, $x=1$ is a clear solution If $x\neq 1$, $nx=\frac{2}{x+1}+\frac{3}{x+2}+\ldots+\frac{n+1}{x+n}$ Once again, $LHS-RHS=0$ $(x-1)\sum_{i=1}^n\frac{x+n+1}{x+n}=0$ As $x\neq 0$ by assumption, and $\sum_{i=1}^n\frac{x+n+1}{x+n}=0$ is not possible as all are positive, so this condition is false. So, $x=1$ is the only solution Q.E.D.

Is my answer correct? Or am I missing something?

the result is correct

This problem also appeared in Latvia 2019 TST for IMO Round 2 Day 1. Solution, which I found during the contest. Note that: $$(nx+\frac{n(n+3)}{2})(n+nx+\frac{n(n+1)}{2})=(nx+\frac{n(n+3)}{2})^2=(n + (x+1) + (x+2)+...+(x+n))(nx^2 +\frac{2^2}{x + 1}+\frac{3^2}{x + 2}+...+\frac{(n + 1)^2}{x + n}= nx + \frac{n(n + 3)}{2})\ge (nx + 2 + 3 + ..(n+1))=(nx+\frac{n(n+3)}{2})^2$$, where we applied Cauchy ineaquality. We conclude that $(n + (x+1) + (x+2)+...+(x+n))(nx^2 +\frac{2^2}{x + 1}+\frac{3^2}{x + 2}+...+\frac{(n + 1)^2}{x + n}= nx + \frac{n(n + 3)}{2})=(nx+\frac{n(n+3)}{2})^2$. This implies ( using when Cauchy inequality achieves equality case): $$\frac{2^2}{(x+1)^2}=\frac{3^2}{(x+2)^2}=...=\frac{(n+1)^2}{(x+n)^2}$$. Therefore $x=1$ and we are done.

Math-wiz wrote: $LHS-RHS=0$ $(x-1)\left[nx-\left(\frac{2}{x+1}+\frac{3}{x+2}+\ldots+\frac{n+1}{x+n}\right)\right]=0$ So, $x=1$ is a clear solution If $x\neq 1$, $nx=\frac{2}{x+1}+\frac{3}{x+2}+\ldots+\frac{n+1}{x+n}$ Once again, $LHS-RHS=0$ $(x-1)\sum_{i=1}^n\frac{x+n+1}{x+n}=0$ As $x\neq 0$ by assumption, and $\sum_{i=1}^n\frac{x+n+1}{x+n}=0$ is not possible as all are positive, so this condition is false. So, $x=1$ is the only solution Q.E.D. sorry but how did you get $(x-1)\left[nx-\left(\frac{2}{x+1}+\frac{3}{x+2}+\ldots+\frac{n+1}{x+n}\right)\right]=0$

Math-wiz wrote: Is my answer correct? Or am I missing something? I also solved it the same way. I think it's correct.