Prove $ 0 \leq \frac{k}{n} \sum_{i = 1}^n \alpha_i - \sum_{i = j}^k \alpha_j$ Given a set of numbers that are non-decreasingly ordered
$$\alpha_1\leq\alpha_2\leq\cdots\leq\alpha_n$$
Prove that 
$$ 0 \leq \frac{k}{n} \sum_{i = 1}^n \alpha_i - \sum_{j = 1}^k \alpha_j$$
for every $k = 1, \dots,n $
The question is related to Proof for A majorizes B , I did not understand the last step in the prof.
 A: You're asked to compare
$$
\frac{\alpha_1+\cdots+\alpha_n}{n}\quad\text{v.s.}\quad\frac{\alpha_1+\cdots+\alpha_k}{k}.
$$
The average on the LHS has at least as many terms as the average on the RHS. Because $\alpha_1\leq\cdots\leq\alpha_n$, adding terms can never decrease the RHS. So the LHS $\geq$ RHS for all $k\in\{1,\ldots,n\}$.
If you prefer to be more rigorous, set $S_k=\frac{1}{k}\sum_{j=1}^k\alpha_j$. Then, for $1\leq k<n$,
\begin{align*}
S_{k+1}-S_k=\frac{1}{k(k+1)}[(\alpha_{k+1}-\alpha_1)+\cdots+(\alpha_{k+1}-\alpha_k)]\geq 0
\end{align*}
which implies $S_1\leq S_2\leq\cdots\leq S_n$. In particular, you have $S_k\leq S_n$ for all $k$, which is your desired inequality.
A: The above statement is true iff 
$$
0\leq k\sum_{i=1}^n\alpha_i-n\sum_{j=1}^k\alpha_j.
$$
Let's eliminate what is common to the two sums: $k\sum_{i=1}^k\alpha_i$.  Therefore, the statement is true iff
$$
0\leq k\sum_{i=k+1}^n\alpha_i-(n-k)\sum_{j=1}^k\alpha_j.
$$
Both sums have the same number of terms, but all of the terms (individually) in the positive sum are greater than the terms of the negative sum.  Hence, the inequality holds.
A: You can write it  $ k \cdot \frac{1}{n} \sum\limits_{i=1}^n \alpha_i  - k \cdot \frac{1}{k} \sum\limits_{i=1}^k \alpha_i   $ and see it as the difference between two average values
