Is the uniform distribution majorized by all sequences? Let $$x = [1/n, 1/n,\dots, 1/n]$$
and $$y = [y_1, \dots, y_n]$$
 be probability vectors on $n$ elements  (i.e., $\sum x_i = \sum y_i = 1$ and $x_i,y_i\geq 0$).
Is it true that $x$ is always majorized by $y$?
My thought: yes. From what I can see on Wikipedia (the Geometry of majorization section), a vector $a$ is majorized by a vector $b$ iff $a$ is within the convex hull of vectors obtained from permuting all the elements of $b$. For probability vectors, the hull will be a subspace of the $n$ dimensional simplex. Intuitively, because $x$ is uniform it will always be at the `centre' of the simplex and therefore within the convex hull. 
I've searched the internet for a while now and have not been able to obtain a reference to such a result. If one has a reference, I would be grateful too!
 A: Let $y_{(1)}, y_{(2)}, \cdots, y_{(n)}$ be a non-increasing order of $y_1, y_2, \cdots, y_n$. For $1 \leq k < n$, let
$$
S_k = \sum_{i=1}^k y_{(i)}
\quad
\text{and}
\quad
T_k = \sum_{i=k+1}^n y_{(i)}
$$
It is easy to see that
$$
S_k + T_k = 1 \tag{$1$}
$$
Moreover, we have
$$
S_k \geq \sum_{i=1}^k y_{(k)} = ky_{(k)} \Rightarrow \frac{S_k}{k} \geq y_{(k)}
\quad
\text{and}
\quad
T_k \leq \sum_{i=k+1}^n y_{(k)} = (n - k)y_{(k)} \Rightarrow \frac{T_k}{n-k} \leq y_{(k)}
$$
That is,
$$
\frac{n-k}{k}S_k \geq T_k \tag{$2$}
$$
Combining $(1)$ and $(2)$, we have
$$
S_k + \frac{n-k}{k}S_k \geq S_k + T_k = 1 \Rightarrow S_k \geq \frac{k}{n}
$$
Therefore, $x$ is majorized by $y$.
A: Let $\{y_i^*\}_{i=1}^n$ be the decreasing (i.e., non-increasing) arrangement of the $y_i$'s. $x$ is majorized by $y$ if and only if $a_k=\frac{1}{k}\sum_{i=1}^ky_i^{*}\geq\frac{1}{n}$ for each $k=1,2,\dots n$. But $a_k$ is non-increasing, as can easily be checked, that is, 
$$a_1\geq a_2\cdots\geq a_n=\frac{1}{n},$$
so the assertion is proved. 
