Prove that $\frac{1}{\sqrt{nx+1}}+\frac{1}{\sqrt{nx+2}}+\ldots+\frac{1}{\sqrt{nx+n}}=\sqrt n$ has a unique positive solution Prove that for each $n\in \mathbb Z$, $n\ge 1$ there is exactly one $x>0$ satisfying 
$$\frac{1}{\sqrt{nx+1}}+\frac{1}{\sqrt{nx+2}}+\ldots+\frac{1}{\sqrt{nx+n}}=\sqrt n$$
Please give me a hint. In fact, after changing the expression, I get the equation:
$$\frac1n\sum\limits_{i=1}^n\dfrac{1}{\sqrt{x+\frac{i}{n}}}=1$$
I don't know what next step I should do. Thank you in advance for your help.

Note that $n=1$ gives $x=0$.
Considering
$$\int_{0}^{1} \frac{dt}{\sqrt{\frac{9}{16}+t}} = 1$$
Since $\frac{1}{\sqrt{\frac{9}{16}+t}}$ is monotonic decreasing, 
the integral is sandwiched by the upper and lower Riemann sums: $U_{n}<U_{n+1}<1<L_{n+1}<L_{n}$
$$\sum_{k=1}^{n} \frac{1}{n\sqrt{\frac{9}{16}+\frac{k}{n}}} < 1 <
  \sum_{k=0}^{n-1} \frac{1}{n\sqrt{\frac{9}{16}+\frac{k}{n}}}$$
$$\sum_{k=1}^{n} \frac{1}{n\sqrt{\frac{9}{16}+\frac{k}{n}}} < 1 <
  \sum_{k=1}^{n} \frac{1}{n\sqrt{\frac{9}{16}-\frac{1}{n}+\frac{k}{n}}}$$
Hence for $n\ge 2$, $\displaystyle \exists x \in \left( \frac{9}{16}-\frac{1}{n}, \frac{9}{16} \right)$ satisfies the equality.
