Prove $E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^k\right]\leq E\left[X_1\left(\frac{X_1+(n-1)\mu}{n}\right)^{k-1}\right]$ $X_1,X_2,\ldots,X_n$ are i.i.d. random variables, $X_1>0$, $E[X_1]=\mu$, $E[X_1^k]<\infty$ for $1<k \leq2$. Prove:
$$
E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^k\right]\leq E\left[X_1\left(\frac{X_1+(n-1)\mu}{n}\right)^{k-1}\right].
$$
I have problem with showing that this inequality really does hold. I know that Jensen inequality should be applied, but I do not know how.
Any help would be great. Thanks!
 A: Writing
$$\left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k = \left( \frac{1}{n} \sum_{j=1}^n X_j \right) \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1}$$
we get
$$\mathbb{E} \left[ \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k \right] = \frac{1}{n} \sum_{j=1}^n \mathbb{E} \left[ X_j \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right].$$
Since the random variables are independent and identically distributed, this implies
$$\mathbb{E} \left[ \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k \right] = \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right]. \tag{1}$$
Using the tower property and the independence of the random variables, we find
$$\begin{align*} \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right] &= \mathbb{E} \left( \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \mid X_1 \right] \right) \\ &= \mathbb{E} \left( X_1 \cdot \mathbb{E} \left[ \left( \frac{x+ \sum_{i=2}^n X_i}{n} \right)^{k-1} \right] \bigg|_{x=X_1} \right).  \tag{2}\end{align*}$$
Since $0<k-1 \leq 1$ it follows from Jensen's inequality that
$$\begin{align*} \mathbb{E} \left[ \left( \frac{x+ \sum_{i=2}^n X_i}{n} \right)^{k-1} \right] &= \left(\mathbb{E} \left[ \left( \frac{x+ \sum_{i=2}^n X_i}{n} \right)^{k-1} \right] \right)^{\frac{k-1}{k-1}} \\ &\leq \left( \mathbb{E} \left[ \frac{x+\sum_{i=2}^n X_i}{n} \right] \right)^{k-1} \\ &= \left( \frac{x+(n-1) \mu}{n} \right)^{k-1}. \tag{3}\end{align*}$$
Combining $(1)$, $(2)$, $(3)$ finishes the proof.
