Hexagonal vs square tiling: comparing total length of edges

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Today I saw this short video: Why do honeybees love hexagons?

The video explains that bees want to store their food (honey) in appropriately sized cells and that wax -the thing that makes the cells- is quite 'expensive' for bees to create. So the bees want to choose an efficient cell shape. What would be the most efficient one? The video continues to explain that a circle is the most efficient shape since it needs the shortest length to enclose a certain size area. But because circles cannot tile a plane, we have to choose a tiling shape.

The main claim of the video is that hexagons (as opposed to squares or triangles) are the most efficient shape for bees to use, because they use less wax to store the same amount of honey. This is definitely true if we are talking about a single cell. A square of 1 unit area has a side of 1 unit length, and a total perimeter of 4 units length, while the equivalent regular hexagon of 1 unit area has a side of $\sqrt\frac{2}{3\sqrt3}\approx 0.62$ and total perimeter length of $6\cdot\sqrt\frac{2}{3\sqrt3}\approx 3.72$. So we need less length (wax) to enclose the same area (honey)

What happens with tiling though? This was not obvious to me despite the video presenting it like so. With tiling, edges are shared. Could this affect the efficiency? I thought about it and I have an answer, I just need other people to double check my reasoning.

First, we have to determine what we are comparing. I think that the tilings need to have the same area cells (whatever their shape) and the same number of cells. This way we are comparing tilings that have the same total area, and the same number of cells. I think that's fair but I am ready to listen to other opinions.

Let's first look at square tiling. Here's an image of an $8\times8$ unit squares:

What is the total length of all edges? It's easy to calculate. We have $9$ horizontal lines of length $8$ and $9$ vertical lines of length $8$. In general, for $n \times n$ unit squares we have $(n+1)n + (n+1)n = 2(n+1)n$ unit lengths.

How does this compare to hexagon tiling? Look at the following image of $8\times8$ unit hexagons:

We again can notice $9$ "zigzag" horizontal lines and $9$ "zigzag" kind of vertical lines. What is the length of each of these lines? As you can see from the image, they span 8 hexagons, and 2 sides of each hexagon making the length of each line $8\times 2\sqrt\frac{2}{3\sqrt3}$. But there is overlap between the horizontal and vertical lines. Let's say we start with the vertical lines, then you can see that we have included half of the horizontal lines as well (except for a line at the boundary). So the total length of edges for a grid of $8\times8$ unit hexagons is $9 \cdot 8\cdot 2h + 9 \cdot 8h + 7h$, where $h=\sqrt\frac{2}{3\sqrt3}$ is the length of the side of the unit hexagon.

In general for a grid of $n\times n$ unit hexagons the total length is:

$$2h\cdot(n+1)n + h\cdot(n+1)n + h\cdot(n-1) = h\cdot(3(n+1)n + n-1) $$

The ratio of the sum of side lengths of the hexagonal tiling over the square tiling is: $$\frac{3h}{2}+ \frac{h(n-1)}{2(n+1)n}$$

So it does depend on $n$. The ratio is always less than $1$ (so hexagonal is always more efficient). It's smallest value is for $n=1$ and its largest value is for $n=2,3$. As $n \rightarrow \infty$ the ratio approaches the value of the ratio for a single cell ($n=1$).

(Note: initially I had found that the ratio does not depend on $n$, which seemed surprising. The problem was that I had missed the term $(n-1)h$ )

My question has three parts:

1. Is my reasoning correct about the main result?
2. Can we do even better in efficiency? For example, what if we tile the hexagons to form a hexagonal bigger shape instead of a rhombus one? Or any other shape. My intuition says we probably can, but will this difference vanish as $n$ gets larger?
3. Is there any other shape that can tile the plane more efficiently in terms of the total length of edges used? Can we prove there isn't one?