<s>[INST] 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!
 [/INST] Another way.
We can use a homogenization.
We need to prove $$\prod_{cyc}\left(a+\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\geq\frac{64}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^3}$$ or
$$\prod_{cyc}(a^2+ab+ac+bc)\geq64a^2b^2c^2$$ or
$$\prod_{cyc}((a+b)(a+c))\geq64a^2b^2c^2$$ or
$$\prod_{cyc}(a+b)\geq8abc$$ or
$$\prod_{cyc}(a+b)\geq8\prod_{cyc}\sqrt{ab}$$ or
$$\prod_{cyc}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\geq8,$$ which is true by AM-GM.
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