Probably, you already got the expected solution (from elementary algebra) in the other answers. So, I will give a different proof (from Multivariable Calculus). Specifically, I will use the Lagrange Multipliers Method to prove the following
Result: Given positive numbers $a$, $b$, $c$ and $d$ with $a+b+c=1$, we have
$$\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)\left(\frac{1}{c+d}+1\right)\geq\left(\frac{1}{\frac{1}{3}+d}+1\right)^3\tag{$*$}$$
Remark 1: Letting $d\to 0$ in $(*)$, we get
$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)\geq\left(\frac{1}{\frac{1}{3}}+1\right)^3=64$$
as you want.
Remark 2: The same approach can be used to prove that for any positive numbers $x_1,...,x_n$ with $x_1+\cdots+x_n=c$ we have
$$\left(\frac{1}{x_1}+1\right)\left(\frac{1}{x_2}+1\right)\cdots\left(\frac{1}{x_n}+1\right)\geq \left(\frac{n}{c}+1\right)^n$$
(See here the case $n=2$.)
Proof of Result: Let $d>0$ and $U=\{(a,b,c);\;a>0,\;b>0,\;c>0\}$. Define $f:U\to\mathbb{R}$ and $\varphi:U\to \mathbb{R}$ by $f(a,b,c)=\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)\left(\frac{1}{c+d}+1\right)$ and $\varphi(a,b,c)=a+b+c$.
Then,
$$\nabla f(a,b,c)=-\left(\tfrac{1}{A^2}\left(\tfrac{1}{B}+1\right)\left(\tfrac{1}{C}+1\right),\tfrac{1}{B^2}\left(\tfrac{1}{A}+1\right)\left(\tfrac{1}{C}+1\right),\tfrac{1}{C^2}\left(\tfrac{1}{A}+1\right)\left(\tfrac{1}{B}+1\right)\right)$$
where $A=a+d$, $B=b+d$ and $C=c+d$, and $\nabla\varphi(a,b,c)=(1,1,1)$.
It follows from the Lagrange Multipliers Method that:
$$(a,b,c)\in \varphi^{-1}(1)\text{ is a critical point of } f\quad\Longleftrightarrow\quad\nabla f(a,b,c)=\lambda \nabla \varphi(a,b,c) \text{ for some } \lambda\in\mathbb{R}$$
$$\Longleftrightarrow \qquad\frac{1}{A^2}\left(\frac{1}{B}+1\right)\left(\frac{1}{C}+1\right)
=\frac{1}{B^2}\left(\frac{1}{A}+1\right)\left(\frac{1}{C}+1\right)
=\frac{1}{C^2}\left(\frac{1}{A}+1\right)\left(\frac{1}{B}+1\right)$$
$$\Longleftrightarrow\qquad \left\{ \begin{align}\frac{1}{A^2}\left(\frac{1}{B}+1\right)
&=\frac{1}{B^2}\left(\frac{1}{A}+1\right)\\
\frac{1}{A^2}\left(\frac{1}{C}+1\right)
&=\frac{1}{C^2}\left(\frac{1}{A}+1\right)\\
\frac{1}{B^2}\left(\frac{1}{C}+1\right)
&=\frac{1}{C^2}\left(\frac{1}{B}+1\right)\end{align}\right.$$
$$\Longleftrightarrow\qquad\left\{\begin{align}
B+B^2&=A+A^2\\
C+C^2&=A+A^2\\
C+C^2&=B+B^2
\end{align}\right.$$
$$\Longleftrightarrow\qquad\left\{\begin{align}
B-A&=A^2-B^2=-(A+B)(B-A)\\
C-A&=A^2-C^2=-(A+C)(C-A)\\
C-B&=B^2-C^2=-(B+C)(C-B)
\end{align}\right.$$
$$\Longleftrightarrow\quad A=B=C$$
$$\Longleftrightarrow\quad a=b=c=\frac{1}{3}$$
So, $q=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ is the unique critical point of $f$ in $\varphi^{-1}(1)$.
The expression of $f(a,b,c)$ defines a continuous function $g$ on $$\overline{\varphi^{-1}(1)}=\{(a,b,c);\;a\geq 0,\;b\geq 0,\;c\geq 0,\;a+b+c=1\}.$$
Since $\overline{\varphi^{-1}(1)}$ is compact, $g$ attains minimum at some point $p=(p_1,p_2,p_3)\in \overline{\varphi^{-1}(1)}$.
If $p\in \overline{\varphi^{-1}(1)}\setminus {\varphi^{-1}(1)}$ then $p_1=0$ or $p_2=0$ or $p_3=0$. Assume $p_1=0$. Then
$$\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)\left(\frac{1}{c+d}+1\right)\geq \left(\frac{1}{d}+1\right)\left(\frac{1}{p_2+d}+1\right)\left(\frac{1}{p_3+d}+1\right)$$
which is a contradiction (because letting $d\to 0$, the LHS goes to some constant and the RHS goes to $\infty$). It follows that $p_1= 0$ is impossible. The same argument shows that $p_2\neq 0$ and $p_3\neq0$. Therefore, $p\in\varphi^{-1}(1)$. Since $g=f$ on $\varphi^{-1}(1)$, we conclude that $f$ attains minimum in $\varphi^{-1}(1)$. Thus, the minimum value of $f$ in $\varphi^{-1}(1)$ is $f(q)$ (because $q$ is the unique critical point of $f$ in $\varphi^{-1}(1)$).
Therefore, for all $(a,b,c)\in\varphi^{-1}(1)$ we have
$$f(a,b,c)\geq f\left(\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3}\right)\tag{#}$$
As $(\text{#})$ is $(*)$, the proof is complete. $\Box $
