comparing $\prod_{j=1}^m \left( \frac1n \sum_{k=1}^n x_k^j\right)^r$ and $\left( \frac1n \sum_{k=1}^n x_k^r\right)^{\binom{m+1}{2}}$ Let $m$ and $n$ be positive integers, and let $r, x_1, x_2,...,x_n$ be positive real numbers. Prove $\prod_{j=1}^m \left( \frac1n \sum_{k=1}^n x_k^j\right)^r\geq \left( \frac1n \sum_{k=1}^n x_k^r\right)^{\binom{m+1}{2}}$ when $r \leq m/2$ while the direction of the inequality changes when $r \geq m$.
I was thinking of using induction but that doesn't seem to give good results. A hint would be greatly appreciated!
 A: To simplify notations, I will use $\mathbb E\left[X^j\right] = \frac1n \sum_{k=1}^n x_k^j$, the two terms of inequality are $$\prod_{j=1}^m \mathbb E\left[X^j\right]^r$$ and $$\mathbb E\left[X^r\right]^{\frac{m(m+1)}{2}} = \mathbb E\left[X^r\right]^{\sum_{j=1}^{m} j}$$

*

*Case 1 : $2r \le m$, let $s = \frac{2r}{m(m+1)}$, then $\frac1s = \frac{m(m+1)}{2r} \ge m+1$. Let $q = \frac{s}{1-sm}=\frac{1}{\frac1s - m} \ge 0$
\begin{align}
\mathbb E\left[X^r\right]^{\frac{m(m+1)}{2}} &= \mathbb E\left[\left(\prod_{j=1}^m X^j\right)^{\frac r{\sum_{j=1}^m j}}\right]^{\frac{m(m+1)}{2}}\\
&= \mathbb E\left[\left(\prod_{j=1}^m X^j\right)^s\right]^{\frac{r}{s}}\\
&= \left(\mathbb E\left[\left(1\times\prod_{j=1}^m X^j\right)^{s}\right]^{\frac1s}\right)^r\\
&\le \left(\mathbb E\left[1^{q}\right]^{\frac1q}\prod_{j=1}^m \mathbb E\left[\left(X^j\right)^1\right]^{\frac11}\right)^r & \text{Holder inequality}.\\
&= \prod_{j=1}^m \mathbb E\left[X^j\right]^r 
\end{align}

*

*Case 2 : $r \ge m$. let $q_j \ge 0$ such that $\frac1{q_j} + \frac jr = 1$
\begin{align}
\prod_{j=1}^m \mathbb E\left[X^j\right]^r &= \prod_{j=1}^m \mathbb E\left[X^j \times 1\right]^r\\
&\le \prod_{j=1}^m \left(\mathbb E\left[\left(X^j\right)^{\frac rj}\right]^{\frac jr} \times \mathbb E\left[1^{q_j}\right]^{q_j}\right)^r & \text{Holder inequality}\\
&= \prod_{j=1}^m \mathbb E\left[X^r\right]^j = \mathbb E\left[X^r\right]^{\frac{m(m+1)}{2}}\\
\end{align}
