Prove $\sum_{k = 1}^n \left(\frac{a_1+a_2+\cdots a_k}{k}\right)^p < \left(\frac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p$ For $p> 1$ and $a_1,a_2,...,a_n$ positive, show that
\begin{equation}
    \sum_{k = 1}^n \left(\frac{a_1+a_2+\cdots a_k}{k}\right)^p < \left(\frac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p
\end{equation}
I was hoping to use the convexity of $f(x) = x^p$, or some induction argument. But I can't seem to figure it out. Some hints would be greatly appreciated!
 A: This is exactly the Hardy's inequality. Here's one possible proof:
Let $A_k=\dfrac{a_{1}+a_{2}+\cdots+a_{k}}{k}$ and define $A_0=0$. From Hölder's inequality, we have
$$
\left(\sum_{k=1}^{n}A_k^p\right)^{\frac{p-1}{p}}\left(\sum_{k=1}^{n}a_k^p\right)^{\frac{1}{p}}\ge\sum_{k=1}^{n}A_k^{p-1}a_k
$$
Hence we need to prove
$$
\sum_{k=1}^{n}A_k^{p-1}a_k\ge\frac{p-1}{p}\sum_{k=1}^{n}A_k^p
$$
Since
$$
\begin{align} LHS&=\sum_{k=1}^{n}A_k^{p-1}(kA_k-(k-1)A_{k-1})\\ &=\sum_{k=1}^{n}kA_k^{p}-\sum_{k=1}^{n}(k-1)A_k^{p-1}A_{k-1}\\ &\ge\sum_{k=1}^{n}kA_k^{p}-\sum_{k=1}^{n}(k-1)\left(\frac{(p-1)A_k^p}{p}+\frac{A_{k-1}^p}{p}\right)(\text{Young's inequality})\\ &=\frac{p-1}{p}\sum_{k=1}^{n-1}A_k^p+(n-\frac{(p-1)(n-1)}{p})A_n^p\\ &\ge\frac{p-1}{p}\sum_{k=1}^{n}A_k^p \end{align}
$$
where Young's inequality is:

(Young's inequality) $$\text{For }x,y,p,q>0,\frac{1}{p}+\frac{1}{q}=1, \text{ we have }\frac{x}{p}+\frac{y}{q}\ge x^\frac{1}{p}y^\frac{1}{q}$$

Thus
$$
\left(\sum_{k=1}^{n}A_k^p\right)^{\frac{p-1}{p}}\left(\sum_{k=1}^{n}a_k^p\right)^{\frac{1}{p}}\ge\sum_{k=1}^{n}A_k^{p-1}a_k\ge\frac{p-1}{p}\sum_{k=1}^{n}A_k^p
$$
Reorganize, we get
$$
\sum_{k=1}^{n}\left(\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}\right)^{p} \leq\left(\frac{p}{p-1}\right)^{p} \sum_{k=1}^{n} a_{k}^{p}
$$
A: By induction, $n=1\implies a_1^p\lt (\dfrac{p}{p-1})^p a_1^p$, since $p\gt 1$.
For $n\gt 1$, may assume $a_{n+1}\ge a_k, \forall k=1,\cdots,n$. Then $\left(\dfrac{a_1+a_2+\cdots a_{n+1}}{n+1}\right)^p\le a_{n+1}^p$
$\sum_{k=1}^{n+1} \left(\dfrac{a_1+a_2+\cdots a_k}{k}\right)^p=\sum_{k=1}^{n} \left(\dfrac{a_1+a_2+\cdots a_k}{k}\right)^p+\left(\dfrac{a_1+a_2+\cdots a_{n+1}}{n+1}\right)^p\lt \left(\dfrac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p+a_{n+1}^p\lt \left(\dfrac{p}{p-1}\right)^p \sum_{k =1}^{n+1} a_k^p$
