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.
 A: Let us consider the function
$$f(x)=\frac1n\sum\limits_{i=1}^n\dfrac{1}{\sqrt{x+\frac{i}{n}}}$$
then $f$ is strictly decreasing and continuous for $x\geq0$ because it is the sum of strictly decreasing and continuous functions.
Moreover $\lim_{x\to +\infty}f(x)=0$ and 
$$f(0)=\frac{1}{\sqrt{n}}\sum_{i=1}^n \frac{1}{\sqrt{i}}\geq \frac{1}{\sqrt{n}}\sum_{i=1}^n \frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n}}\cdot \frac{n}{\sqrt{n}}=1.$$
Hence for any $n\in\mathbb{N}^+$, there is a unique $x_0\geq 0$ such that $f(x_0)=1$.
A: 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.
A: Let us call $f(x)$ the LHS of the equation.
$f(x)$ is defined for  $x>-1/n$
$f'(x)=\sum_{i=1}^n \dfrac{(-3/2)}{(nx+i)^{3/2}}$ is $<0$.
Thus $f$ is a strictly decreasing function with a vertical asymptote with equation $x=-1/n$ and the $x$-axis as horizontal asymptote because $f(x)$ tends to 0 when $x$ tends to $+\infty$. 
Therefore $f:(-1/n,\infty) \rightarrow (0,\infty)$ is bijective.
Consequently, there is a unique $x_0$ such that $f(x_0)=n$.
Moreover, one can say that  $x_0 \geq 0$ because 
$$\sum_{i=1}^n \dfrac{1}{\sqrt{i}} \geq \sqrt{n} \ \ \ (1)$$
(See this reference)
Remark: as remarked by @Ng Chung Tak , one has equality in (1) for $n=1$.
A: n=1 would lead to x=0, however the posted constraint on the solution was X > 0.
In that case there exists no solution. Shouldn't the problem be rephrased?
