Question

If \( a>0, b>0, c>0 \) and \( a, b, c \) are distinct, then \( (a + b)(b + c)(c + a) \) is greater than:

• A. \( 2(a + b + c) \)
• B. \( 3(a + b + c) \)
• C. \( 6abc \)
• D. \( 8abc \)

Hint:

Use the AM-GM inequality to compare products of sums for positive distinct values. This is helpful in many problems involving inequalities.

Answer 1

The Correct Option is D

We are given \( a>0, b>0, c>0 \) and the condition that \( a, b, c \) are distinct. The goal is to find which of the following expressions \( (a + b)(b + c)(c + a) \) is greater than. 
Step 1: Use AM-GM inequality
We know that the Arithmetic Mean is greater than or equal to the Geometric Mean: \[ AM \geq GM. \] Applying this to the terms \( a + b, b + c, c + a \), we get: \[ (a + b)(b + c)(c + a) \geq 8abc. \] Thus, the correct answer is \( 8abc \), which matches option (D). 
 

Answer 2

To determine which of the given options is less than \( (a + b)(b + c)(c + a) \), let's analyze the expression:  

Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), we know that for any positive numbers \( x \) and \( y \), \(\frac{x + y}{2} \geq \sqrt{xy} \). Applying this to the terms, we have: 
1. \( a + b \geq 2\sqrt{ab} \) 
2. \( b + c \geq 2\sqrt{bc} \) 
3. \( c + a \geq 2\sqrt{ca} \) 

Multiplying these inequalities, we get: 
\[ (a + b)(b + c)(c + a) \geq 8\sqrt{a^2b^2c^2} = 8abc \] 
Since \( a, b, c \) are distinct positive numbers, equality does not hold, so: 
\[ (a + b)(b + c)(c + a) > 8abc \] 
Evaluating the options provided:

Options	Inequality Check
\(2(a + b + c)\)	\((a + b)(b + c)(c + a)\) cannot be compared directly with this form.
\(3(a + b + c)\)	\((a + b)(b + c)(c + a)\) cannot be compared directly with this form.
\(6abc\)	\((a + b)(b + c)(c + a) > 8abc \Rightarrow (a + b)(b + c)(c + a) > 6abc\)
\(8abc\)	\((a + b)(b + c)(c + a) > 8abc\)

✅ Correct Option: \(8abc\) is less than \((a + b)(b + c)(c + a)\)