Find The Minimum Value of the quantity Find the minimum value of the quantity $$\frac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc}$$,where $$a,b,c>0$$ and $$ a,b,c\in R $$are positive real numbers.

Hint: $$\frac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc} = \left(3+a+\frac1a \right)\left(3+b+\frac1b \right)\left(3+c+\frac1c \right)$$
Now can you show that $x + \dfrac1x$ has a minimum of $2$ for positive real $x$?
