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. \(6 abc\)
• D. \(8 abc\)

Hint:

In competitive exams, if you see products like \((a+b)(b+c)(c+a)\) with positive numbers, the AM-GM inequality is usually the first tool to use. If the numbers are not specified as distinct, the answer might be \(\geq 8abc\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for an inequality condition for the product of sums of three distinct positive real numbers \(a, b, c\).

Step 2: Key Formula or Approach:
The Arithmetic Mean-Geometric Mean (AM-GM) Inequality states that for positive real numbers, \(\frac{x+y}{2} \geq \sqrt{xy}\).
Equality holds if and only if \(x = y\). Since \(a, b, c\) are distinct, the sum of any two will be strictly greater than twice the square root of their product.

Step 3: Detailed Explanation:
Applying AM-GM to pairs of distinct numbers \(a, b, c\):
1. For \(a\) and \(b\): \(\frac{a+b}{2} > \sqrt{ab} \Rightarrow a+b > 2\sqrt{ab}\).
2. For \(b\) and \(c\): \(\frac{b+c}{2} > \sqrt{bc} \Rightarrow b+c > 2\sqrt{bc}\).
3. For \(c\) and \(a\): \(\frac{c+a}{2} > \sqrt{ca} \Rightarrow c+a > 2\sqrt{ca}\).
Multiply these three inequalities together (valid since all terms are positive):
\[ (a+b)(b+c)(c+a) > (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca}) \] \[ (a+b)(b+c)(c+a) > 8 \sqrt{a^2 b^2 c^2} \] \[ (a+b)(b+c)(c+a) > 8 abc \]

Step 4: Final Answer:
The expression is greater than \(8 abc\).