Well, I was confronted with a wonderful problem like this:
If $\{a_n\}$ is a positive sequence such that $\displaystyle \sum_{n=1}^\infty \frac1{a_n}<\infty$. Let $\displaystyle S_n=\sum_{k=1}^na_k$. Then for any $\alpha\geqslant0$, $$ \sum_{n=1}^{\infty}{\left( \frac{na_n}{S_n} \right) ^{\alpha}\frac{1}{a_n}}<\infty $$ Moreover,my friend guesses that the following inequality may be correct $$ \sum_{n=1}^{\infty}{\left( \frac{na_n}{S_n} \right) ^{\alpha}\frac{1}{a_n}}\leqslant2^{\alpha}\sum_{n=1}^{\infty}{\frac{1}{a_n}} $$ And the constant $2^\alpha$ is the best choice. For $\alpha=0$ and $\alpha=1$, it's not hard to prove. But for $\alpha\in\mathbb R_+$, I don't know how to deal with it.
Or the equivalent inequality of integral like this:
$f(x)\geqslant0$,Let$\displaystyle F(x)=\int_0^xf(t)\mathrm dt$. Prove that $$ \int_0^{\infty}{\left( \frac{xf\left( x \right)}{F\left( x \right)} \right) ^{\alpha}\frac{\text{d}x}{f\left( x \right)}}\leqslant 2^{\alpha}\int_0^{\infty}{\frac{\text{d}x}{f\left( x \right)}} $$ The inquality may be right only for$0\leqslant\alpha\leqslant2$.
Any advice will be appreciated.Thank you!
By Knopp's inequality (see pages 109+ of my notes) we have $$ \sum_{n=1}^{N}\frac{n}{S_n}< 2\sum_{n=1}^{N}\frac{1}{a_n}\tag{1}$$ and by Holder's inequality, for any $\alpha\in(0,1)$ $$ \sum_{n=1}^{N}\left(\frac{n}{S_n}\right)^{\alpha}\left(\frac{1}{a_n}\right)^{1-\alpha}\leq \left(\sum_{n=1}^{N}\frac{n}{S_n}\right)^\alpha\left(\sum_{n=1}^{N}\frac{1}{a_n}\right)^{1-\alpha}\stackrel{(1)}{<}2^\alpha\sum_{n=1}^{N}\frac{1}{a_n}\tag{2} $$ as claimed. The optimality of such inequality should not be difficult to check by considering $a_n=n$ for any $n\leq M$ and $a_n=n(n+1-M)$ for any $n\geq M$. The discrete version implies the continuous version by approximations through simple functions.