If a, b, c are positive integers,then (a + b) | vedclass

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(B) Since the Arithmetic Mean $(AM)$ is greater than the Geometric Mean $(GM)$ for positive real numbers,we have:$\frac{a + b}{2} > \sqrt{ab}$$\frac{b

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$> 8abc$

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(B) Since the Arithmetic Mean $(AM)$ is greater than the Geometric Mean $(GM)$ for positive real numbers,we have:$\frac{a + b}{2} > \sqrt{ab}$$\frac{b + c}{2} > \sqrt{bc}$$\frac{c + a}{2} > \sqrt{ca}$Multiplying these three inequalities together,we get:$\left(\frac{a + b}{2}\right) \left(\frac{b + c}{2}\right) \left(\frac{c + a}{2}\right) > \sqrt{ab} \cdot \sqrt{bc} \cdot \sqrt{ca}$$\frac{(a + b)(b + c)(c + a)}{8} > \sqrt{a^2 b^2 c^2}$$\frac{(a + b)(b + c)(c + a)}{8} > abc$Therefore,$(a + b)(b + c)(c + a) > 8abc$.

If $a, b, c$ are positive integers,then $(a + b)(b + c)(c + a)$ is

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