A proof that isn't very clever but does the job: let
$$f(a,b,c)=ab\left(a+b+c-\tfrac ab-\tfrac bc-\tfrac ca\right)$$
which is chosen so that $f(a_n,b_n,c_n)\ge 0$.

The idea is:

- Reduce to the $abc=1$ case by defining $g(a,b)=f(a,b,\frac{1}{ab})$ and proving that $f(a_n,b_n,c_n)-g(a_n,b_n)\to 0$.
- Prove that $g$ is nonpositive and that $g(a_n,b_n)\to 0$ implies $(a_n,b_n)\to (1,1)$.
- Conclude that $(a_n,b_n,c_n)\to (1,1,1)$.

Proof:

Since the problem is invariant under cyclic permutation of $a_n,b_n,c_n$ for any single $n$, we can assume that $c_n$ is the maximum. We get
$$c_n/a_n\le a_n+b_n+c_n\le 3c_n\\
1/a_n\le 3$$
This implies $\limsup b_n c_n\le 3$, so that
$$\limsup a_n^2 b_n^3\le \limsup c_n^2 b_n^{5/2} c_n^{1/2}\le 3^{5/2}$$
So:
$$a_n^2 b_n^3\left(\tfrac{1}{a_nb_nc_n}-1\right)+\tfrac{1}{a_n}(a_nb_nc_n-1)\to 0$$
and therefore
$$f(a_n,b_n,c_n)-g(a_n,b_n)\to 0$$

A straightforward calculation shows that $g(a,b)$ is the cubic polynomial $$-(ab)^3+(ab)^2+a^3b-a^3+a-1$$ with discriminant in $a$
$$-(b-1)^2(b+1)(23b^3+5b^2-27b+23)$$
so that, on $D=[0,+\infty)^2$, $g(a,b)$ is always non-zero when $b\ne 1$, and when $b=1$ it is non-zero only when $a=1$. Therefore because $g(a,b)$ is negative at $(0,0)$ it is everywhere negative except at $(a,b)=(1,1)$.

Furthermore the $\sup$ of $g$ over $D$ minus any neighborhood of $(1,1)$ is negative because $D$ is closed and $g\le -1+\varepsilon$ in a neighborhood of $\infty$. As a consequence, if $g(a_n,b_n)\to 0$ then $(a_n,b_n)\to (1,1)$.

Because $f(a_n,b_n,c_n)\ge 0$ and $g$ is nonpositive, $g(a_n,b_n)\to 0$, so that $(a_n,b_n)\to (1,1)$ and therefore:
$$\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=\lim_{n\to\infty} c_n=1$$
which is actually a stronger form of the theorem.