It is given that $\sum_{r=1}^{n} \dfrac{7r+4}{r(r+1)(r+2)} = \dfrac{9}{2} - \dfrac{2}{n+1} - \dfrac{5}{n+2}.$

1. Rewrite the general term of the summation by factorising its denominator.

$\dfrac{7r-3}{r^3-r} = \dfrac{7r-3}{r(r^2-1)} = \dfrac{7r-3}{(r-1)r(r+1)}$

2. Perform an index substitution to match the form of the given summation.

$\text{Let } k = r-1. \text{ Then } r = k+1.$
$\text{When } r = 7, k = 6.$
$\text{When } r = 2n, k = 2n-1.$
$\text{The term becomes } \dfrac{7(k+1)-3}{k(k+1)(k+2)} = \dfrac{7k+4}{k(k+1)(k+2)}.$
$\text{Therefore, } \sum_{r=7}^{2n} \dfrac{7r-3}{r^3-r} = \sum_{k=6}^{2n-1} \dfrac{7k+4}{k(k+1)(k+2)}.$

3. Express the sum starting from $k=6$ as the difference of two sums starting from $k=1$.

$\sum_{k=6}^{2n-1} \dfrac{7k+4}{k(k+1)(k+2)} = \sum_{k=1}^{2n-1} \dfrac{7k+4}{k(k+1)(k+2)} - \sum_{k=1}^{5} \dfrac{7k+4}{k(k+1)(k+2)}$

4. Apply the given summation formula to both parts and simplify to obtain the final answer.

$\text{Using the given formula } \sum_{r=1}^{n} \dfrac{7r+4}{r(r+1)(r+2)} = \dfrac{9}{2} - \dfrac{2}{n+1} - \dfrac{5}{n+2}:$
$\text{For } n = 2n-1: \quad \sum_{k=1}^{2n-1} \dots = \dfrac{9}{2} - \dfrac{2}{(2n-1)+1} - \dfrac{5}{(2n-1)+2} = \dfrac{9}{2} - \dfrac{1}{n} - \dfrac{5}{2n+1}$
$\text{For } n = 5: \quad \sum_{k=1}^{5} \dots = \dfrac{9}{2} - \dfrac{2}{5+1} - \dfrac{5}{5+2} = \dfrac{9}{2} - \dfrac{1}{3} - \dfrac{5}{7}$
$\text{Substituting back:}$
$\left( \dfrac{9}{2} - \dfrac{1}{n} - \dfrac{5}{2n+1} \right) - \left( \dfrac{9}{2} - \dfrac{1}{3} - \dfrac{5}{7} \right) = -\dfrac{1}{n} - \dfrac{5}{2n+1} + \dfrac{1}{3} + \dfrac{5}{7} = \dfrac{22}{21} - \dfrac{1}{n} - \dfrac{5}{2n+1}$

1. Find the difference between the left-hand side and the right-hand side of the inequality.

$(r+1)^3 - r(r+1)(r+2)$
$= (r+1)[(r+1)^2 - r(r+2)]$
$= (r+1)(r^2 + 2r + 1 - r^2 - 2r)$
$= (r+1)(1)$
$= r+1$

2. Use the condition that $r$ is a positive integer to deduce the inequality.

Since $r$ is a positive integer ($r \ge 1$), we have $r+1 > 0$.
Therefore,
$(r+1)^3 - r(r+1)(r+2) > 0$
$(r+1)^3 > r(r+1)(r+2)$

1. Deduce an inequality for the terms of the series using the result from (b).

From part (b), we have $(r+1)^3 > r(r+1)(r+2)$.
Since $r \ge 1$, both sides are positive. Taking the reciprocal reverses the inequality:
$\frac{1}{(r+1)^3} < \frac{1}{r(r+1)(r+2)}$
Since $7r+4 > 0$ for $r \ge 1$, multiplying both sides gives:
$\frac{7r+4}{(r+1)^3} < \frac{7r+4}{r(r+1)(r+2)}$

2. Apply the summation to both sides and substitute the given series result.

Summing both sides from $r=1$ to $n$:
$\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \sum_{r=1}^{n} \frac{7r+4}{r(r+1)(r+2)}$
Substituting the expression given in the question stem:
$\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2}$

3. Conclude the proof by evaluating the signs of the subtracted terms.

For all positive integers $n \ge 1$, the terms $\frac{2}{n+1}$ and $\frac{5}{n+2}$ are strictly positive.
Therefore, subtracting positive quantities from $\frac{9}{2}$ results in a value strictly less than $\frac{9}{2}$:
$\frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2} < \frac{9}{2}$
Hence, it follows that:
$\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \frac{9}{2}$