已知 $\displaystyle\sum_{r=1}^{n} \frac{7r+4}{r(r+1)(r+2)} = \frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2}。$

1. Factorise the denominator and match the summand's structure to the given identity.

Factorising the denominator of the given summation term:
$r^3 - r = r(r^2 - 1) = (r-1)r(r+1)$
So the term is $\frac{7r-3}{(r-1)r(r+1)}$.

Notice that if we substitute $k = r - 1$ (which means $r = k + 1$), the term becomes:
$\frac{7(k+1)-3}{k(k+1)(k+2)} = \frac{7k+4}{k(k+1)(k+2)}$
This matches the exact expression in the given sum.

2. Apply the substitution to rewrite the summation limits.

Adjusting the summation limits for the substitution $k = r - 1$:
When the lower limit $r = 7$, $k = 6$.
When the upper limit $r = 2n$, $k = 2n - 1$.

Thus, the summation can be rewritten as:
$\sum_{r=7}^{2n} \frac{7r-3}{r^3-r} = \sum_{k=6}^{2n-1} \frac{7k+4}{k(k+1)(k+2)}$

3. Express the new summation as the difference of two series starting from $k=1$.

To use the given formula which starts from the first term, we split the sum:
$\sum_{k=6}^{2n-1} \frac{7k+4}{k(k+1)(k+2)} = \sum_{k=1}^{2n-1} \frac{7k+4}{k(k+1)(k+2)} - \sum_{k=1}^{5} \frac{7k+4}{k(k+1)(k+2)}$

4. Evaluate the difference using the given formula and simplify.

We are given $\displaystyle\sum_{r=1}^{n} \frac{7r+4}{r(r+1)(r+2)} = \frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2}$.

For the upper sum, substitute $n = 2n - 1$:
$\sum_{k=1}^{2n-1} \frac{7k+4}{k(k+1)(k+2)} = \frac{9}{2} - \frac{2}{(2n-1)+1} - \frac{5}{(2n-1)+2} = \frac{9}{2} - \frac{1}{n} - \frac{5}{2n+1}$

For the subtracted sum, substitute $n = 5$:
$\sum_{k=1}^{5} \frac{7k+4}{k(k+1)(k+2)} = \frac{9}{2} - \frac{2}{5+1} - \frac{5}{5+2} = \frac{9}{2} - \frac{1}{3} - \frac{5}{7} = \frac{9}{2} - \frac{22}{21}$

Subtracting the two evaluates to our final expression:
$\left( \frac{9}{2} - \frac{1}{n} - \frac{5}{2n+1} \right) - \left( \frac{9}{2} - \frac{22}{21} \right) = \frac{22}{21} - \frac{1}{n} - \frac{5}{2n+1}$

1. Find the difference between the left hand side and the right hand side.

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

2. Conclude the inequality based on the values $r$ can take.

$\text{Since } r \text{ is a positive integer, } r \ge 1.$
$\therefore r + 1 > 0$
$\text{Since the difference is positive, } (r+1)^3 > r(r+1)(r+2). \quad \text{(shown)}$

1. Form an inequality for the terms of the sum using the result from part (b).

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

2. Sum the inequalities from $r=1$ to $n$ and substitute the given series sum.

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 sum expression given in the main 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 observing the subtracted terms are positive.

For any positive integer $n \ge 1$, both $\frac{2}{n+1}$ and $\frac{5}{n+2}$ are strictly positive.
Therefore, subtracting these positive fractions from $\frac{9}{2}$ results in a value strictly less than $\frac{9}{2}$:
$\frac{9}{2} - \left( \frac{2}{n+1} + \frac{5}{n+2} \right) < \frac{9}{2}$
Hence, we can conclude that:
$\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \frac{9}{2}$