Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows:

Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that
$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \ge 64$$

and the proof provided was the following:
Note that $$ abc \le (\frac{a+b+c}{3})^3 = \frac{1}{27} \tag{1}$$
by AM-GM inequality. Then
$$(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc} \tag{2}$$
$$\ge 1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{(abc)^2}} +\frac
{1}{abc} \tag{3}$$
$$=(1+\frac{1}{\sqrt[3]{abc}})^3 \ge 4^3 \tag{4}$$
Steps 1 and 2 are easy for me to understand, but if someone could help me with steps 3 and 4 ,I would be very thankful.

I have any solution.
Note that
$a+1=a+a+b+c\geq 4\sqrt[3]{a^2bc} \tag{1}$
by AM-GM
Analog we get that
$b+1=b+a+b+c\geq 4\sqrt[3]{ab^2c} \tag{2}$
and
$c+1=c+a+b+c\geq 4\sqrt[3]{abc^2} \tag{3}$
So
$$(a+1)(b+1)(c+1)\geq 64abc\Longleftrightarrow (\dfrac{a+1}{a})(\dfrac{b+1}{b})(\dfrac{c+1}{c})\geq 64$$
The equality will hold by $a=b=c=1$.
Proved
