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

$A.M.\ge G.M.$

How i can achieve this?


1 Answer 1


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.

  • 1
    $\begingroup$ @vikiiii: You are welcome. $\endgroup$
    – Aryabhata
    Mar 6, 2012 at 7:20

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