a vector inequality and combinatorics related question This question is a similar restatement of this question which has been recently closed.

Let
$$A=\{\ (x,y,z)\in\mathbb{N}^3\ |\ 0\leq x,y,z\leq7\}$$
and
$$B\subset A \text{ with } |B|\geq49.$$
Show that there exists two distinct vectors $\ x,\ y \in B$ such that
  $x \leq y$,
  where the inequality is defined element-wise.

Trying to get a contradiction, construct B such that the inequality does not hold for any distinct pair. Then, $(0,0,0)$ and $(7,7,7)$ are definitely not in B. If, in the question, A were defined as:
$A=\{\ (x,y,z)\in\mathbb{N}^3\ |\ 0\leq x,y,z\leq1\}$
and if $|B|\geq 4$ then at least one of $(1,0,0), (0,1,0)$ and $(0,0,1)$ would be in B, say $(1,0,0)\in B$. This implies $(0,0,0),(1,1,0),(1,0,1),(1,1,1) \notin B$. So,
$B= \{\ (1,0,0),(0,1,0),(0,0,1),(0,1,1)\ \}$.
A contradiction!

I have tried to utilise the pigeonhole principle and use geometric intuition ( the convex cone in $\mathbb{R}^3$) but I only managed to end up with a rather complex summation formula having $49\pm1$ terms with the inclusion- exclusion principle. I am guessing that there is an easier and more general approach.
 A: Turns out it is quite easy (this is a bit informal, but if you feel the need I'm sure you can make it rigorous very easily):
We consider every z-slice of the cube separately: the slice $z=0$, $z=1$, etc. We first consider the slices in an unordered fashion, place elements on each slice separately, and then stack the slice on top of each other. We are going to place as many elements on the slices as possible, while ensuring that no $2$ are comparable. 
It's not hard to see that on each of these slices there can be at most $8$ elements from $B$ (placed on the diagonal). So we place our first $8$ elements from $B$ on the main-diagonal of one of the slices. 
Clearly, once we placed $8$ elements on the diagonal of one of the slices, we can now place at most $7$ elements on any other slice (by placing them on the length-7 next to the main diagonal) (because elements directly above each other are comparable, we cannot use the main diagonal again). There are $2$ such length-7 diagonals, so we can place $7$ elements on the length-7 diagonals of $2$ of the slices.
Clearly now any other slice can have at most $6$ elements on it, again on one of the length-6 diagonals.
We continue in this way, place: $5,6,7,8,7,6,5,4$ elements on the slices. Note the total of $48$. If we use this order of slice from top to bottom we obtain a valid stacking of the slices, in which no $2$ elements are comparable. It's not hard to convince yourself that it's not possible to place any more elements on the slices while still being able to stack them together s.t. no elements are comparable.
Further, it's now easy to generalise: for an $n\times n\times n$ cube:
$$\begin{align}
&n=0: 0\\
&n=1: 1\\
&n=2: 2+1=3\\
&n=3: 3+2+2=7\\
&n=4: 4+3+3+2=12\\
&n=5: 5+4+4+3+3=19\\
&n=6: 6+5+5+4+4+3=27\\
&n=7: 7+6+6+5+5+4+4=37\\
&n=8: 8+7+7+6+6+5+5+4=48\\
&n=9: 9+8+8+7+7+6+6+5+5=61\\
&\text{etc.}
\end{align}$$
Altough the pattern it clear, a closed form is $\frac{(n+1)(n+2)}{2}+\lfloor\frac{n^2}{4}\rfloor$ . See A077043. You can take a look at some of the descriptions to see some equivalent problems to the one you asked (equivalent in the sense that they have the same answers as a function of $n$).

Edit: in case my explaination did not really make it clear, here are some picture for various $n$:

With the mathematica code:
MakeGrid[n_, {x_, y_}, z_] := Array[
  If[x - #1 == -(y - #2),
    Cuboid[{{#1, #2, z}, {#1 + 1, #2 + 1, z + 1}}],
    ## &[]
    ] &
  , {n, n}]
n = 8;
l = Table[Ceiling[(n + 1)/2] + i, {i, 0, n - 1}];
B = Table[MakeGrid[n, {1, l[[i]]}, n - i], {i, 1, Length[l]}];
Print[Length[Flatten[B]]];
Graphics3D[{Red, B}, Axes -> True, AxesLabel -> {x, y, z}]

A: You can try to cover $A$ with 48 chains. (Chain is a set where any two elements are comparable.) Then, the statement follows easily from the pigeonhole's principle since at least two elements of $B$ must be in the same chain, thus comparable.
Note that if the question is correct, the chain cover of size 48 'must' exist by Dilworth's theorem.
