Proof for a question regarding order of $n$-tuples of real numbers Question as an image
Let $x = (x_1, x_2, x_3,..., x_n)$ and $y= (y_1, y_2, y_3,..., y_n),$ where $x_1, x_2, x_3,..., x_n ,y_1, y_2, y_3,..., y_n$ are real numbers.
We write $x > y$ if, for some $k$, $1 \le k \le (n-1)$,
$x_1 = y_1$, $x_2 = y_2$, $x_3 = y_3$, ...,  $x_k= y_k$ but $x_{k+1} > y_{k+1}$.
Show that for $u = ( u_1,..., u_n)$, $v = ( v_1,..., v_n)$, $w = ( w_1,..., w_n)$, $z = ( z_1,..., z_n)$, if $u > v$ and $w > z$, then $(u+w) > (v+z)$.
I have lots of doubts. Does $(u+w)$ mean that I add each corresponding $i$th element of the sets and thus have still $n$ elements in $(u+w)$ or does it mean that I just put together all the elements from each of the set and then arrange it ascendingly to have $2n$ elements?
If it is the latter, then I think we can just prove that elements $( u_{k+1} +  w_{n+1} ) > (v_{k+1} +  z_{n+1}),$ hence get the sets also in that inequality for some $m \le k+n$. I am not sure what topic this problem which might have helped. It is from a general problems book for a competitive exam.
Can anyone please help me with this problem? Thanks.
I am having trouble with Mathjax, so I have also attached an image with the right formatting. 
 A: Let $1 \le k_1 \le n - 1$ be such that $u_i = v_i$ for all $1 \le i \le k_1$ but $u_{k_1 + 1} \gt v_{k_1 + 1}$. Also, let $1 \le k_2 \le n - 1$ be such that $w_i = z_i$ for all $1 \le i \le k_2$ but $w_{k_2 + 1} \gt z_{k_2 + 1}$. Next, we have the following $3$ cases to consider:
$$k_1 \lt k_2 \tag{1}\label{eq1}$$
$$k_1 = k_2 \tag{2}\label{eq2}$$
$$k_1 \gt k_2 \tag{3}\label{eq3}$$
To make somewhat easier to discuss, let $a = u + w$, so $a_i = u_i + w_i$ for $1 \le i \le n$, and $b = v + z$, so $b_i = v_i + z_i$ for $1 \le i \le n$. With \eqref{eq1}, we have that $a_i = b_i$ for all $1 \le i \le k_1$. However, for $i = k_1 + 1$, we have that $u_{k_1 + 1} \gt v_{k_1 + 1}$, but $w_{k_1 + 1} = z_{k_1 + 1}$, so $a_{k_1 + 1} \gt b_{k_1 + 1}$. A similar situation occurs for \eqref{eq3}, but with $k_1$ and $k_2$ switched around.
Thus, the final case to consider is \eqref{eq2}, where I let $k = k_1 = k_2$. In this case, we still have that $a_i = b_i$ for all $1 \le i \le k$. Also, both $u_{k + 1} \gt v_{k + 1}$ and $w_{k + 1} \gt z_{k + 1}$, so $a_{k + 1} \gt b_{k + 1}$.
In all $3$ cases, we have that $a \gt b$, i.e., $\left(u + w\right) \gt \left(v + z\right)$.
