Given a < b < 0, which of the following inequalities is true?

We can subtract $ab$ from both sides to get $a^2 - ab > 0$.

$a(a - b) > 0$.

• Since $a < 0$, $a$ is a negative number.
• Since $a < b$, if we subtract $b$ from both sides, we get $a - b < 0$. So, $(a - b)$ is also a negative number.

(negative number) $\times$ (negative number) = (positive number).
So, $a(a - b)$ is indeed positive, meaning $a(a - b) > 0$ is true.
Therefore, $a^2 > ab$ is true.

Example: Let $a = -3$ and $b = -2$.
$a^2 = (-3)^2 = 9$
$ab = (-3)(-2) = 6$
Is $9 > 6$? Yes, it is.

B. $-\frac{1}{a} > -\frac{1}{b}$

Since $a$ and $b$ are both negative, taking the reciprocal of both sides reverses the inequality sign:

This also reverses the inequality sign:

This contradicts the statement in option B. Therefore, option B is false.

Example: Let $a = -3$ and $b = -2$.
$-\frac{1}{a} = -\frac{1}{-3} = \frac{1}{3}$
$-\frac{1}{b} = -\frac{1}{-2} = \frac{1}{2}$
Is $\frac{1}{3} > \frac{1}{2}$? No, $\frac{1}{3} < \frac{1}{2}$.

C. $-\frac{1}{a} < \frac{1}{b}$

• Since $a < 0$, $\frac{1}{a}$ is negative. Therefore, $-\frac{1}{a}$ is positive.
• Since $b < 0$, $\frac{1}{b}$ is negative.

A positive number is always greater than a negative number. So, $-\frac{1}{a} > \frac{1}{b}$.
This contradicts the statement in option C. Therefore, option C is false.

Example: Let $a = -3$ and $b = -2$.
$-\frac{1}{a} = \frac{1}{3}$
$\frac{1}{b} = -\frac{1}{2}$
Is $\frac{1}{3} < -\frac{1}{2}$? No, $\frac{1}{3} > -\frac{1}{2}$.

D. $ab < b^2$

We can subtract $b^2$ from both sides to get $ab - b^2 < 0$.

$b(a - b) < 0$.

• Since $b < 0$, $b$ is a negative number.
• Since $a < b$, it means $a - b$ is a negative number.

(negative number) $\times$ (negative number) = (positive number).
So, $b(a - b)$ is positive, meaning $b(a - b) > 0$.
This contradicts the statement in option D. Therefore, option D is false.

Example: Let $a = -3$ and $b = -2$.
$ab = (-3)(-2) = 6$
$b^2 = (-2)^2 = 4$
Is $6 < 4$? No, $6 > 4$.

Based on the analysis, only option A is true.

The final answer is $\boxed{\text{A}}$