Let $f(x)=\dfrac{x}{\sqrt{x+1}}$.

1. Apply a substitution to transform the integral into standard integrable forms and integrate.

Let $u = x+1$. Then $\frac{du}{dx} = 1 \implies dx = du$, and $x = u-1$.
When $x = 0$, $u = 1$.
When $x = 1$, $u = 2$.
$\int_{0}^{1} \frac{x}{\sqrt{x+1}}\,dx = \int_{1}^{2} \frac{u-1}{\sqrt{u}}\,du$
$= \int_{1}^{2} \left( u^{\frac{1}{2}} - u^{-\frac{1}{2}} \right)\,du$
$= \left[ \frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}} \right]_{1}^{2}$

2. Evaluate the definite integral using the upper and lower limits, and simplify to the given exact form.

$= \left( \frac{2}{3}(2)^{\frac{3}{2}} - 2(2)^{\frac{1}{2}} \right) - \left( \frac{2}{3}(1)^{\frac{3}{2}} - 2(1)^{\frac{1}{2}} \right)$
$= \left( \frac{2}{3}(2\sqrt{2}) - 2\sqrt{2} \right) - \left( \frac{2}{3} - 2 \right)$
$= \left( \frac{4}{3}\sqrt{2} - \frac{6}{3}\sqrt{2} \right) - \left( -\frac{4}{3} \right)$
$= -\frac{2}{3}\sqrt{2} + \frac{4}{3}$
$= \frac{2}{3}\left(2 - \sqrt{2}\right) \quad \text{(shown)}$

1. Express the area of the $k$-th rectangle.

The $k$-th rectangle has width $\frac{1}{n}$ and its right edge is at $x = \frac{k}{n}$. The height of this rectangle is the function value at this right endpoint:
$f\left(\frac{k}{n}\right) = \frac{\frac{k}{n}}{\sqrt{\frac{k}{n}+1}} = \frac{\frac{k}{n}}{\sqrt{\frac{k+n}{n}}} = \frac{k}{n} \cdot \frac{\sqrt{n}}{\sqrt{k+n}} = \frac{k}{\sqrt{n}\sqrt{n+k}}$

The area of the $k$-th rectangle is:
$\text{Area}_k = \text{width} \times \text{height} = \frac{1}{n} \times \frac{k}{\sqrt{n}\sqrt{n+k}} = \frac{k}{n\sqrt{n}\sqrt{n+k}}$

2. Sum the areas of all $n$ rectangles to find the total area.

The total area of the $n$ rectangles is the sum of their individual areas from $k=1$ to $n$:
$\text{Total Area} = \sum_{k=1}^{n} \frac{k}{n\sqrt{n}\sqrt{n+k}} = \frac{1}{n\sqrt{n}} \left\{ \frac{1}{\sqrt{n+1}} + \frac{2}{\sqrt{n+2}} + \cdots + \frac{n}{\sqrt{2n}} \right\}$

3. Deduce the inequality by comparing the total area to the definite integral.

Since $f(x)$ is an increasing function for $x \ge 0$, the right-endpoint rectangles strictly overestimate the exact area under the curve. Thus, the total area of the rectangles is strictly greater than the definite integral of $f(x)$ from $x=0$ to $x=1$:
$\text{Total Area} > \int_{0}^{1} f(x)\,dx$

Substituting the total area expression and using the exact value of the integral from part (i):
$\frac{1}{n\sqrt{n}} \left\{ \frac{1}{\sqrt{n+1}} + \frac{2}{\sqrt{n+2}} + \cdots + \frac{n}{\sqrt{2n}} \right\} > \frac{2}{3}\bigl(2-\sqrt{2}\bigr)$