Maybe an integral inequality in discrete form?

Question

If $\alpha$ is a real number in $(0,1)$ and $a_1,a_2, \dots ,a_N \geq 0$, prove that $$\sum\limits_{n=1}^{N} \left(\sum\limits_{k=n}^{N}\frac{a_k}{k} \right)^\alpha \geq \alpha^\alpha \sum\limits_{n=1}^{N} a_n^\alpha.$$

This looks like a reverse of Karamata's inequality, and probably has some higher algebra backgrounds. I hope someone can help me solve it.

Thanks!

I've tried the norm inequality and the power mean inequality,but it failed — flower417477, Nov 16 '25 at 12:06
This looks like it could use Cauchy-Schwarz or Minkowski inequalities, maybe paired with summation by parts. — Maxence1402, Nov 18 '25 at 08:44
This is one of Copson's inequalities presented in 1928 in his paper. In 1987, Bennett extended it in his own article. — Bulat, Nov 26 '25 at 18:27

score 1 · Answer 1 · edited Nov 22 '25 at 12:13

After several days I finally find a solution to this question.

This actually can be proven like the Hardy's inequality

Let $A_k = \sum\limits_{k=n}^{N}\frac{a_k}{k} $

Then $a_n = n(A_n-A_{n+1})$ , where $A_{N+1}=0$

By Holder's inequality $\sum\limits_{n=1}^{N} a_n^\alpha \leq \left(\sum\limits_{n=1}^{N} A_n^\alpha\right)^{1-\alpha} \left(\sum\limits_{n=1}^{N} \frac{a_n}{A_n^{1-\alpha}}\right)^{\alpha}$

So we only need that $\alpha \sum\limits_{n=1}^{N} \frac{a_n}{A_n^{1-\alpha}} \leq \sum\limits_{n=1}^{N} A_n^\alpha$

which is $\alpha \sum\limits_{n=1}^{N} \frac{n(A_n-A_{n+1})}{A_n^{1-\alpha}} \leq \sum\limits_{n=1}^{N} A_n^\alpha$

As Young's inequality gives $\alpha \frac{A_{n+1}}{A_n^{1-\alpha}} + (1-\alpha)A_n^\alpha \geq A_{n+1}^\alpha$

So $LHS \leq \sum\limits_{n=1}^{N} (n\alpha A_n^\alpha - n (A_{n+1}^\alpha-(1-\alpha)A_n^\alpha) ) = \sum\limits_{n=1}^{N} n(A_n^\alpha - A_{n+1} ^\alpha) = RHS$

So we're done.

Maybe an integral inequality in discrete form?

1 Answers1