在不使用图形计算器的情况下，解不等式 $\dfrac{x+4}{x+1} \ge \dfrac{x+2}{2}$。 由此，解 $1 + \dfrac{3}{e^x + 1} \ge \dfrac{e^x + 2}{2}$。

1. Rearrange the inequality to make one side zero and combine the terms into a single algebraic fraction.

$\frac{x+4}{x+1} - \frac{x+2}{2} \ge 0$
$\frac{2(x+4) - (x+1)(x+2)}{2(x+1)} \ge 0$

2. Expand and simplify the numerator, then factorise it.

$\frac{2x + 8 - (x^2 + 3x + 2)}{2(x+1)} \ge 0$
$\frac{-x^2 - x + 6}{2(x+1)} \ge 0$
$\frac{-(x^2 + x - 6)}{2(x+1)} \ge 0$
$\frac{-(x+3)(x-2)}{2(x+1)} \ge 0$

3. Rewrite the inequality with a positive leading coefficient and identify the critical values.

Multiply both sides by $-1$ and reverse the inequality sign (the constant $2$ in the denominator can also be divided out as it is positive):
$\frac{(x+3)(x-2)}{x+1} \le 0$

The critical values are found by setting the numerator and denominator to zero:
$x = -3$, $x = 2$, and $x = -1$.

4. Perform a sign test to find the intervals that satisfy the inequality.

Using a number line or sign test for the expression $\frac{(x+3)(x-2)}{x+1}$:

• For $x < -3$, expression is negative.
• For $-3 < x < -1$, expression is positive.
• For $-1 < x < 2$, expression is negative.
• For $x > 2$, expression is positive.

We require the expression to be $\le 0$. The critical value $x = -1$ makes the denominator zero, so it must be excluded ($x \neq -1$).

The solution set is $x \le -3 \text{ or } -1 < x \le 2$.

1. Solve the rational inequality from part (a) as it is required for part (b).

For part (a), solving $\frac{x+4}{x+1} \ge \frac{x+2}{2}$:
$\frac{2(x+4) - (x+1)(x+2)}{2(x+1)} \ge 0$
$\frac{2x+8 - (x^2 + 3x + 2)}{2(x+1)} \ge 0$
$\frac{-x^2 - x + 6}{2(x+1)} \ge 0 \implies \frac{(x+3)(x-2)}{x+1} \le 0$
Using a sign test with critical values $-3$, $-1$, and $2$, the solution to (a) is:
$x \le -3 \text{ or } -1 < x \le 2$

2. Rewrite the inequality in (b) to match the form in (a) and apply the previous solution.

Rewrite the left-hand side of the inequality in (b):
$1 + \frac{3}{e^x + 1} = \frac{e^x + 1 + 3}{e^x + 1} = \frac{e^x + 4}{e^x + 1}$
The inequality becomes $\frac{e^x + 4}{e^x + 1} \ge \frac{e^x + 2}{2}$, which matches the exact form in (a) with $x$ replaced by $e^x$.
Substituting $e^x$ into the solution from (a):
$e^x \le -3 \text{ or } -1 < e^x \le 2$

3. Solve the resulting exponential inequalities.

Since $e^x > 0$ for all real $x$:

• The inequality $e^x \le -3$ has no real solutions.
• The condition $-1 < e^x$ is always satisfied, leaving $e^x \le 2$.

Taking the natural logarithm on both sides:
$x \le \ln 2$