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Find the minimum value of the quantity where a , b , c are real positive numbers.

$$\left(\frac{a^2 + 3a + 1}{a}\right) \left(\frac{b^2 +3b + 1}{b}\right)\left(\frac{c^2 + 3c + 1}{c}\right) $$

I think the to get the answer we need to use

A.M.  >= G.M 

How i can achieve this?

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1 Answer 1

up vote 2 down vote accepted

You can use the fact that $x + \frac{1}{x} \ge 2$, which can be proved using $\text{AM} \ge \text{GM}$, or just completing the square.

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Thanks. Got it. – vikiiii Mar 6 '12 at 7:19
@vikiiii: You are welcome. – Aryabhata Mar 6 '12 at 7:20

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