I am looking for a formal proof shows that for any $x\&y \geq 0$ and $\alpha > 1$,
\begin{equation} \frac{1}{n}\sum_{i=1}^{n}x_i ^\alpha \geq \frac{1}{n}\sum_{i=1}^{n}y_i ^\alpha \end{equation}
where $STD(x_1...x_i) \geq STD(y_1...y_i)$ and $\sum_{i=1}^{n}x_i = \sum_{i=1}^{n}y_i$
I think I should use Generalized Mean Inequality but I don't know how because in GMI for any $s \geq 0$ and $\alpha > 1$,
\begin{equation} (\frac{1}{n}\sum_{i=1}^{n}s_i^\alpha) \geq (\frac{1}{n}\sum_{i=1}^{n}s_i)^\alpha \end{equation}
It can derives from the fact that since RHS has $STD = 0$ (we can say that we have $n$ numbers with the same value) but I think it is not a proof for this problem since I want to put it in an academic paper and also this inequality only shows a special case when $STD = 0$
Thanks