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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!

Math_Day
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2 Answers2

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

FFjet
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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 $\left(\dfrac{a_1+a_2+\cdots a_{n+1}}{n+1}\right)^p\le a_{n+1}^p$

$\sum_{k=1}^{n+1} \left(\dfrac{a_1+a_2+\cdots a_k}{k}\right)^p=\sum_{k=1}^{n} \left(\dfrac{a_1+a_2+\cdots a_k}{k}\right)^p+\left(\dfrac{a_1+a_2+\cdots a_{n+1}}{n+1}\right)^p\lt \left(\dfrac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p+a_{n+1}^p\lt \left(\dfrac{p}{p-1}\right)^p \sum_{k =1}^{n+1} a_k^p$

Divide1918
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  • I don't think you can assume $a_{n+1}\geq a_n$. The LHS of this inequality isn't symmetric among $a_1,...,a_n$. You'll have to show that among $n!$ permutations of $a_1,...,a_n$, it is true that $a_{n+1}\geq a_n$ maximizes the LHS – Math_Day Dec 22 '20 at 17:41
  • @Will Given any n numbers $a_1,\cdots , a_n\in\Bbb R$ they must be ordered, i.e. $a_{\sigma(1)}\le \cdots\le a_{\sigma(n)}$ for some permutation $\sigma\in S_n$. Re-index the variables $a_j$ by replacing $a_j$ by $a_{\sigma(j)}$. – Divide1918 Dec 23 '20 at 03:12
  • @Will Another way to look at it is that if you are given $(n+1)$ numbers $a_j$, you can pick out the largest of them all and call it $a_{n+1]$, and when you remove that, the rest n numbers satisfy the inequality by induction hypothesis. – Divide1918 Dec 23 '20 at 03:15