Question

please anyone answer this question​

Answer 1

Plz pardon if its wrong, I am not sure about it correctness

Answer 2

AM>=GM

$\\ a + \frac{1}{a} > = 2$

now

$\frac{ {a }^{2} + 3a + 1}{a} \\ = a + \frac{1}{a } + 3$

$> = 2 + 3$

$> = 5$

similarly

$\frac{ {b}^{2} + 3b + 1}{b} > = 5 \\ and \\ \frac{ {c}^{2} + 3c + 1}{c} > = 5$

multiply to get

$\frac{( {a}^{2} + 3a + 1)( {b}^{2} + 3b + 1)( {c}^{2} + 3c + 1) }{abc} > = 125$

minimum value is 125