Question

( b + c )( c + a )( a + b ) is greater than

• A. $abc$
• B. $2abc$
• C. $4abc$
• D. $8abc$

Answer 1

The correct option is D

$8abc$

Explanation for correct options:

Step 1: Declaring the theorem:

We know, A.M = $\frac{a+b}{2}$

G.M = $\sqrt{ab}$

For two positive numbers, A.M $\ge$ G.M

Step 2: Comparing the given equation with the theorem:

For the given problem,

$\Rightarrow \frac{a+b}{2}\ge \sqrt{ab}$.. (1)

$\Rightarrow \frac{b+c}{2}\ge \sqrt{bc}$.. (2)

$\Rightarrow \frac{a+c}{2}\ge \sqrt{ac}$.. (3)

Step 3: Solving the equations:

Now, multiplying equations (1, 2, 3),

$\Rightarrow \frac{a+b}{2}\times \frac{b+c}{2}\times \frac{a+c}{2}\ge \sqrt{ab}\times \sqrt{bc}\times \sqrt{ac}$

$\Rightarrow \frac{(a+b)(b+c)(a+c)}{8}\ge \sqrt{ab}\times \sqrt{bc}\times \sqrt{ac}$

$\Rightarrow \frac{(a+b)(b+c)(a+c)}{8}\ge \sqrt{a^2\times b^2\times c^2}$

$\Rightarrow \frac{(a+b)(b+c)(a+c)}{8}\ge abc$

$\Rightarrow (a+b)(b+c)(a+c)\ge abc$

Therefore, the expression $(a+b)(b+c)(a+c)$ is greater than $8abc$

Hence, Option (D) is correct.