2
$\begingroup$

I am struggling with the following task:

Let$$1<p<\infty , n\in \mathbb{N}, a_1\geqslant 0,\dots,a_n\geqslant 0.$$ Prove that $$\sum_{k=1}^n \Bigl(\frac{a_1+\dots+a_k}{k}\Bigr)^p \le \frac{p}{p-1}\Biggl(\sum_{k=1}^n \Bigl(\frac{a_1+\dots+a_k}{k}\Bigr)^{p-1}a_k\Biggr)$$ and $$\Biggl(\sum_{k=1}^n \Bigl(\frac{a_1+\dots+a_k}{k}\Bigr)^p\Biggr)^{1/p} \le \frac{p}{p-1} \Biggl(\sum_{k=1}^n a_k^p \Biggr)^{1/p}.$$

My attempt: Let $f(z)=z^p$, then \begin{align*} \sum_{k=1}^n \Bigl(\frac{a_1+\dots+a_k}{k}\Bigr)^p &=\sum_{k=1}^n f\Bigl(\frac{a_1+\dots+a_k}{k}\Bigr)\\ &=\sum_{k=1}^n \int_0^{\frac{a_1+\dots+a_k}{k}} f'(t)dt\\ &=\sum_{k=1}^n \Bigl(\int_0^{\frac{a_1}{k}} f'(t)dt+\dots+\int_{\frac{a_1+\dots+a_{k-1}}{k}}^{\frac{a_1+\dots+a_k}{k}} f'(t)dt \Bigr)\\ &\le \sum_{k=1}^n \Biggl(\frac{a_1}k f'\Bigl(\frac{a_1}k\Bigr)+\frac{a_2}kf'\Bigl(\frac{a_1+a_2}k\Bigr)+\dots+\frac{a_k}kf'\Bigl(\frac{a_1+\dots+a_k}k\Bigr)\Biggr)\\ &=\sum_{k=1}^n \frac pk\Biggl(a_1\Bigl(\frac{a_1}k\Bigr)^{p-1}+a_2\Bigl(\frac{a_1+a_2}k\Bigr)^{p-1}+\dots+a_k\Bigl(\frac{a_1+\dots+a_k}k\Bigr)^{p-1}\Biggr)\\ &=\sum_{k=1}^n \frac pk\Biggl(\sum_{j=1}^k a_j \Bigl(\frac{a_1+\dots+a_j}{k}\Bigr)^{p-1}\Biggr), \end{align*}

but I'm not sure how to proceed. Could anyone help, please? Thanks

$\endgroup$
1
1
$\begingroup$

I'm not sure if it is possible to prove the inequality following your steps, so if you will excuse me, I will just post my solution.

First inequality:

Let $S_k = (a_1 + \dots + a_k )/k$ for each valid $k$, and define $S_0 = 0$. We shall prove that $$ \left( 1 + \frac{k}{p-1}\right) S_k^p \leq \frac{p}{p-1} S_k^{p-1} a_k + \frac{k-1}{p-1} S_{k-1}^p $$ for all $1\leq k\leq n$. If this is true, summing the inequalities for $k = 1,2,\dots,n$, you will get the desired result.

To prove the inequality above, substitute $a_k = kS_k - (k-1)S_{k-1}$, then it reduces to $$ p S_k^{p-1} (S_{k-1}-S_k) \leq S_{k-1}^p - S_k^p $$ This can be easily proved by dividing the cases by whether $S_k\geq S_{k-1}$ or not.

Second Inequality is Hardy's Inequlity

$\endgroup$
1
  • $\begingroup$ Thank you so much for the help! $\endgroup$
    – Chaos
    Jun 9 '19 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.