In virtue of this result: Prove that $\sum_{k=1}^n \frac{2k+1}{a_1+a_2+...+a_k}<4\sum_{k=1}^n\frac1{a_k}.$ it is possible to state that, if $$\sum_{n=1}^{+\infty}\frac{1}{a_n}$$ is a converging series with positive terms, $$\sum_{n=1}^{+\infty}\frac{n}{a_1+\ldots+a_n}<2\sum_{n=1}^{+\infty}\frac{1}{a_n},$$ and this is exactly the statement of the Hardy's inequality for $p=-1$.

(1) Is the constant $2$ in the RHS the best possible constant?

(2) Does the integral analogue holds? I.e., is it true that, if $f$ is a positive function belonging to $L^1(\mathbb{R}^+)$, $$\int_{0}^{+\infty}\left(\frac{1}{x}\int_{0}^{x}\frac{dy}{f(y)}\right)^{-1}\,dx<2\int_{0}^{+\infty}f(x)\,dx\;?$$

(3) Does the Hardy-type inequality with negative exponent $$\int_{0}^{+\infty}\left(\frac{1}{x}\int_{0}^{x}\frac{dy}{f(y)^p}\right)^{-1/p}\,dx< C_p \int_{0}^{+\infty}f(x)\,dx\;?$$ holds for any $p\geq 1$? If so, what are the best possible constants $C_p$?


I managed to prove many things through the following inequality.

For any $p\geq 1$ and for every $a,b,\alpha,\beta>0$ we have: $$\left(\frac{(\alpha+\beta)^{p+1}}{a^p+b^p}\right)^{1/p}\leq \left(\frac{\alpha^{p+1}}{a^p}\right)^{1/p}+ \left(\frac{\beta^{p+1}}{b^p}\right)^{1/p}.$$ If we set $b/a=x$, it is sufficient to prove that the minimum of the function $f:\mathbb{R}^+\to \mathbb{R}^+$ defined by: $$ f(x) = \alpha^{\frac{p+1}{p}}(1+x^p)^{1/p} + \beta^{\frac{p+1}{p}}(1+x^{-p})^{1/p} $$ is exactly $(\alpha+\beta)^{\frac{p+1}{p}}$. To do that, it is sufficient to consider that $f'(x)$ has a unique zero in $$ x = \left(\frac{\beta}{\alpha}\right)^{1/p}. $$

Using this inequality, I managed to show that for any real number $p\geq 1$ there exists a constant $C_p\in\mathbb{R}^+$ such that, if $a_1,\ldots,a_N$ are positive real numbers, $$\sum_{n=1}^{N}\left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p}< C_p\sum_{n=1}^{N}\frac{1}{a_n}$$ holds. I prove that there exists a positive increasing function $f:\mathbb{N}_0\to\mathbb{R}^+$ for which: $$(\diamondsuit)\left(\frac{f(N)}{\sum_{n=1}^N a_n^p}\right)^{1/p}+\sum_{n=1}^{N}\left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p} \leq \frac{C_p}{a_N}+ \left(\frac{f(N-1)}{\sum_{n=1}^{N-1} a_n^p}\right)^{1/p}+\sum_{n=1}^{N-1}\left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p},$$ such that, by induction, we have: $$ \left(\frac{f(N)}{\sum_{n=1}^N a_n^p}\right)^{1/p}+\sum_{n=1}^{N}\left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p} \leq \frac{1+f(1)^{1/p}}{a_1}+\sum_{n=2}^{N}\frac{C_p}{a_n}.$$ In order to $(\diamondsuit)$ imply the discrete "reverse Hardy" inequality it is sufficient to have $f(1)\leq(C_p-1)^p$ and: $$ \forall N\geq 2,\qquad \left(\frac{f(N)}{\sum_{n=1}^N a_n^p}\right)^{1/p}+\left(\frac{N}{\sum_{n=1}^N a_n^p}\right)^{1/p} \leq \frac{C_p}{a_n}+\left(\frac{f(N-1)}{\sum_{n=1}^{N-1} a_n^p}\right)^{1/p}.$$ In virtue of the initial inequality, if we have $f(N)^{1/p}+N^{1/p}\geq C_p^{p/(p+1)}$, then: $$ \frac{f(N)^{1/p}+N^{1/p}}{\left(\sum_{n=1}^N a_n^p\right)^{1/p}} \leq \frac{C_p}{a_N}+\frac{\left(\left(f(N)^{1/p}+N^{1/p}\right)^{\frac{p}{p+1}}-C_p^{\frac{p}{p+1}}\right)^{\frac{p+1}{p}}}{\left(\sum_{n=1}^{N-1}a_n^p\right)^{1/p}},$$ so it suffices to find a $f$ such that: $$ \left(f(N)^{1/p}+N^{1/p}\right)^{\frac{p}{p+1}}\leq f(N-1)^{\frac{1}{p+1}}+C_p^{\frac{p}{p+1}}.$$ Now we take $C_p = (1+p)^{\frac{1}{p}}$, since this is the best possible constant in the "reverse Hardy" inequality if $a_n=n$, then we take $f(N) = k\cdot N^{p+1}$; the previous inequality become: $$ (\spadesuit)\quad k^{\frac{1}{p+1}} N \left(1+\frac{1}{N k^{1/p}}\right)^{\frac{p}{p+1}}\leq k^{\frac{1}{p+1}}(N-1)+ (1+p)^{\frac{1}{p+1}}.$$ In virtue of the Bernoulli inequality we have: $$ \left(1+\frac{1}{N k^{1/p}}\right)^{\frac{p}{p+1}} \leq 1 + \frac{p}{N (p+1) k^{1/p}}, $$ so, if we find a positive $k$ such that: $$ (\heartsuit)\qquad \frac{p}{p+1}\,k^{-\frac{1}{p(p+1)}}+k^{\frac{1}{p+1}}\leq C_p^{\frac{p}{p+1}}=(p+1)^{\frac{1}{p+1}}$$ the inequality $(\spadesuit)$ is fulfilled. By studying the stationary points of $g(x)=A x^{-\alpha}+ x^{\beta}$ it is quite simple to prove that, with the choice $$ k = (p+1)^{-p} $$ $(\heartsuit)$ holds as an equality. The last thing is to verify that, with the choice $f(N) = \frac{N^{p+1}}{(p+1)^p}$ we have $f(1)\leq (C_p-1)^p$, or $C_p\geq 1+\frac{1}{p+1}$, or: $$ (p+1)^{\frac{1}{p}} \geq 1+\frac{1}{p+1}. $$ By multiplying both sides by $(p+1)$ we have that the inequality is equivalent to: $$ (p+1)^{\frac{p+1}{p}} \geq p+2, $$ that is a consequence of the Bernoulli inequality, since: $$ (p+1)^{\frac{p+1}{p}} \geq 1 + \frac{p+1}{p}\cdot p = p+2. $$

This proves that for any $p\geq 1$ and for any sequence $a_1,\ldots,a_N$ of positive real numbers we have:

$$ \frac{N^{\frac{p+1}{p}}}{(p+1)\left(a_1^p+\ldots+a_N^p\right)^{1/p}}+\sum_{n=1}^N \left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p} \leq (1+p)^{\frac{1}{p}}\sum_{n=1}^{N}\frac{1}{a_n},$$ that is a substantial extension of the discrete Hardy inequality to negative exponents, with an optimal constant, too.

Once proven the discrete version, proving the integral version should be quite straightforward, now.


(2) is also a direct consequence of the Godunova's inequality $$ \int_0^{+\infty} \phi\left(\frac{1}{x}\int_0^x g(t)\,dt\right)\frac{dx}{x}\leq\int_0^{+\infty}\phi(g(x))\frac{dx}{x}$$ which holds for any positive convex function $\phi$ over $\mathbb{R}^+$. It suffices to consider $\phi(x)=\frac{1}{x}$ then take the change of variable $y=x^2$ to have: $$\int_0^{+\infty}\frac{dy}{\frac{1}{y}\int_0^y g(z)\,dz}\leq 2\int_0^{+\infty}\frac{dy}{g(y)}.$$


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