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?
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.