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!
Best Answer
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:
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 $$ \sum_{k=1}^{n}\left(\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}\right)^{p} \leq\left(\frac{p}{p-1}\right)^{p} \sum_{k=1}^{n} a_{k}^{p} $$