Finding minimum value of $\mu$ in cubic $x^3-\lambda x^2+\mu x-6=0$

Question

If $\lambda,\mu$ are the real number such that, $x^3-\lambda x^2+\mu x-6=0$ has its real roots and positive, then the minimum value of $\mu$ is?

My attempts:

As it has real and positive roots its derivate too have real positive roots, i.e. $3x^2-2\lambda x+\mu=0$ applying $D\geq0\implies \lambda^2\geq3\mu>0$. I don't know how to use that $6$, may as the product of roots of a cubic, but where?

How to proceed, please help.

Answer 1

Hint

Assume the roots are $a,b,c.$ Then

$$x^3-\lambda x^2+\mu x-6=(x-a)(x-b)(x-c),$$ from where

$$abc=6, ab+ac+bc=\mu.$$ Since $a,b,c>0$ we have that (AM-GM inequality)

$$\mu=ab+ac+bc\ge 3\sqrt[3]{a^2b^2c^2}=3\sqrt[3]{36}.$$

Is it possible to achieve the minimum value?