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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 $C_p$?

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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.

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(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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