To prove the inequality $a \cdot b + 9 < 3 \cdot (a + b)$ given the condition $a < 3 < b$, we will proceed with the following steps:

Step 1: Start with the given inequality:

We need to prove: $a \cdot b + 9 < 3 \cdot (a + b)$

Step 2: Expand the right-hand side:

The right-hand side can be expanded as: $3 \cdot (a + b) = 3a + 3b$

So, the original inequality becomes: $a \cdot b + 9 < 3a + 3b$

Step 3: Rearrange the inequality:

We can rearrange the terms to group $a \cdot b$ and $3a + 3b$: $a \cdot b - 3a - 3b + 9 < 0$

Step 4: Rewrite the inequality:

Rearrange the left-hand side as: $a \cdot b - 3a - 3b + 9 = (a - 3)(b - 3)$

Thus, the inequality becomes: $(a - 3)(b - 3) < 0$

Step 5: Analyze the inequality $(a - 3)(b - 3) < 0$:

Since $a < 3$ and $b > 3$, we have:

• $a - 3 < 0$ because $a$ is less than 3.
• $b - 3 > 0$ because $b$ is greater than 3.

Therefore, the product $(a - 3)(b - 3)$ is negative: $(a - 3)(b - 3) < 0$

This confirms that: $a \cdot b + 9 < 3 \cdot (a + b)$

Hence, the original inequality is proved.

Would you like more details on any part of this proof?

Related Questions:

1. How would the proof change if the conditions on $a$ and $b$ were different?
2. Can you prove a similar inequality with different coefficients instead of 3?
3. What would happen if both $a$ and $b$ were less than 3?
4. How would the inequality be affected if one variable equaled 3?
5. Can this proof be extended to other mathematical inequalities?

Tip:

When solving inequalities, try factoring expressions and understanding the sign of each factor in the given conditions.