Proving that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1\implies (a+1)(b+1)(c+1)\geq 64$ where $a,b,c>0$. I am trying to solve the following problem:

Let $a,b,c>0$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that:
$$(a+1)(b+1)(c+1)\geq 64$$

So far, I have gotten that by AM-GM, $(a+1)\geq 2\sqrt{a}$, $(b+1)\geq 2\sqrt{b}$ and $(c+1)\geq 2\sqrt{c}$ so:
$$(a+1)(b+1)(c+1)\geq 8\sqrt{abc} \tag{1}$$
Then using AM-GM on $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, we get that:
\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\
1 & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\
\frac{1}{3} & \geq \frac{1}{\sqrt[3]{abc}} \\
\frac{1}{27} & \geq \frac{1}{abc} \\
abc & > 27
\end{align}
Substituting this into $(1)$, we get:
$$(a+1)(b+1)(c+1)\geq 8\sqrt{27}$$
So I clearly went wrong somewhere, though I'm not sure why. It would be best if you could provide a solution using AM-GM or Muirhead's Inequality. The question seems so simple, where equality happens at $a=b=c=3$, though I can't prove this. Thank you in advance for your solutions and hints!

we use $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$,with $x+y+z=1$
we have to prove $(1+x)(1+y)(1+z)\ge 64xyz$
But $$1+x=x+y+z+x\ge 4{(x^2yz)}^{1/4}$$ similarly  can you do it for $1+y$ and $1+z$ ?.
Multiplying the above results you get dezired inequality.
