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.

If a < 3 < b then prove that a*b + 9 < 3*(a + b)

This detailed guide explains how to prove the inequality a * b + 9 < 3 * (a + b) under the condition that a < 3 < b. It covers important algebraic techniques like factoring and rearranging terms, making it perfect for high school students (Grades 9-12).

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