About the Polya-Knopp-like inequality $\sum_{k=1}^{n}\frac{k^2}{a^2_{1}+\cdots+a^2_{k}}\le\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)^2$

I was inspired by other question I came out with the inequality:

let $a_{i}>0,i=1,2,\cdots,n$ Prove that $$\sum_{k=1}^{n}\dfrac{k^2}{a^2_{1}+\cdots+a^2_{k}}\le\left(\dfrac{1}{a_{1}}+\cdots+\dfrac{1}{a_{n}}\right)^2\tag{1}$$ and I believe that (1) follows, by some way, from Carleman's inequality and Hardy's inequality but I did not manage to prove it.

case $n=2$,$$\Longleftrightarrow \dfrac{1}{x^2}+\dfrac{4}{x^2+y^2}\le\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\Longleftrightarrow \dfrac{4}{x^2+y^2}\le\dfrac{1}{y^2}+\dfrac{2}{xy}\Longleftrightarrow (x+2y)(x^2+y^2)\ge 4xy^2$$ This is clear hold,because $x^2+y^2\ge 2xy,x+2y\ge 2y$

Solution 1:

We will use an approach similar to this answer. First, a few inequalities. $$ \begin{align} \sum_{j=1}^kj^{1/2} &=\sum_{j=1}^k\left(\int_{j-1/2}^{j+1/2}x\,\mathrm{d}x\right)^{\!\!1/2}\\ &\ge\sum_{j=1}^k\int_{j-1/2}^{j+1/2}x^{1/2}\,\mathrm{d}x\\ &=\int_{1/2}^{k+1/2}x^{1/2}\,\mathrm{d}x\\[6pt] &=\frac23\left[(k+1/2)^{3/2}-(1/2)^{3/2}\right]\tag{1} \end{align} $$

For $k\gt0$ $$ \begin{align} &\frac{\mathrm{d}}{\mathrm{d}k}\left[(k+1/2)^{3/2}-(1/2)^{3/2}-k^{2/3}(k+1/2)^{5/6}\right]\\ &=\left[9k^{1/3}(k+1/2)^{2/3}-4(k+1/2)-5k\right]\frac{k^{-1/3}(k+1/2)^{-1/6}}6 \end{align} $$ which has the same sign as $$ 729k(k+1/2)^2-(9k+2)^3=\frac14\left[972k^2+297k-32\right] $$ which is positive for $k\ge1/2$. Therefore, since $1\gt(1/2)^{3/2}+(1/2)^{2/3}$, for $k\ge1/2$ $$ (k+1/2)^{3/2}-(1/2)^{3/2}\gt k^{2/3}(k+1/2)^{5/6}\tag{2} $$

$$ \begin{align} \frac23\left(k^{-3/2}-(k+1)^{-3/2}\right) &=\int_k^{k+1}x^{-5/2}\,\mathrm{d}x\\ &\ge\left(\int_k^{k+1}x\,\mathrm{d}x\right)^{-5/2}\\[7pt] &=(k+1/2)^{-5/2}\tag{3} \end{align} $$

Combining these three inequalities yields $$ \begin{align} \frac1{k^2}\left(\sum_{j=1}^kj^{1/2}\right)^{\!\!3} &\ge\frac8{27k^2}\left[(k+1/2)^{3/2}-(1/2)^{3/2}\right]^3\tag{4a}\\ &\ge\frac8{27}(k+1/2)^{5/2}\tag{4b}\\[6pt] &\ge\frac49\frac1{k^{-3/2}-(k+1)^{-3/2}}\tag{4c} \end{align} $$

Hölder's Inequality implies that $$ \begin{align} \frac1{k^2}\left(\sum_{j=1}^ka_j^2\right)\left(\sum_{j=1}^k\frac{j^{3/4}}{a_j}\right)^{\!\!2} &\ge\frac1{k^2}\left(\sum_{j=1}^kj^{1/2}\right)^{\!\!3}\\[4pt] &=\sigma_k\tag{5} \end{align} $$ Therefore, $$ \begin{align} \sum_{k=1}^n\raise{4pt}{\frac{k^2}{\sum\limits_{j=1}^ka_j^2}} &\le\sum_{k=1}^n\frac1{\sigma(k)}\left(\sum_{j=1}^k\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &=\sup_{\|c\|_2=1}\left(\sum_{k=1}^n\frac{c_k}{\sigma_k^{1/2}}\sum_{j=1}^k\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &=\sup_{\|c\|_2=1}\left(\sum_{j=1}^n\sum_{k=j}^n\frac{c_k}{\sigma_k^{1/2}}\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &\le\left(\sum_{j=1}^n\left(\sum_{k=j}^n\frac1{\sigma_k}\right)^{\!\!1/2}\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &\le\left(\sum_{j=1}^n\frac32j^{-3/4}\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &=\frac94\left(\sum_{j=1}^n\frac1{a_j}\right)^{\!\!2}\tag{6} \end{align} $$

Solution 2:

Following Omran Kouba's approach in a linked question, from the Holder inequality we have: $$\sum_{j=1}^{k} j^2 \leq \sqrt[3]{\sum_{j=1}^{j}j^3 a_j}\cdot \sqrt[3]{\sum_{j=1}^{j}j^3 a_j}\cdot\sqrt[3]{\sum_{j=1}^{k}\frac{1}{a_j^2}}\tag{0}$$ from which it follows that: $$\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq \frac{27}{k(k+1)^3(k+1/2)^3}\left(\sum_{j=1}^{k}j^3 a_j\right)^2\leq\frac{9}{2}\left(\frac{1}{k^6}-\frac{1}{(k+1)^6}\right)\left(\sum_{j=1}^{k}j^3 a_j\right)^2 $$ and by Cauchy-Schwarz inequality: $$\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq\frac{9}{2}\left(\frac{1}{k^6}-\frac{1}{(k+1)^6}\right)\left(\sum_{j=1}^{k}a_j\right)\left(\sum_{j=1}^{k}j^6 a_j\right).\tag{1}$$ If we set $S_k\triangleq \frac{1}{k^6}\sum_{j=1}^{k-1}j^6 a_j$ and $S_1=0$ we have: $$\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq \frac{9}{2}(S_k-S_{k+1}+a_k)\sum_{j=1}^{k}a_j\leq \frac{9}{2} a_k \sum_{j=1}^{k}a_j.\tag{2}$$ Now we set $A_k=\sum_{j=1}^{k}a_j.$ From the previous line: $$\sum_{k=1}^{n}\frac{k^2}{\sum_{j=1}^{k}\frac{1}{a_j^2}}\leq\frac{9}{2}\sum_{k=1}^{n} a_k A_k \tag{3}$$ and from summation by parts: $$ \sum_{k=1}^{n}a_k A_k = A_n^2 - \sum_{k=2}^{n}A_{k-1} a_{k}=A_n^2-\sum_{k=1}^{n}A_k a_k+\sum_{k=1}^{n}a_k^2 $$ hence: $$ 2\sum_{k=1}^{n}a_k A_k = \left(\sum_{j=1}^{n}a_j\right)^2+\sum_{j=1}^{n}a_j^2 $$ proves your inequality up to a multiplicative factor $\color{red}{\frac{9}{2}}$.

Edit: If in line $(0)$ we replace the LHS with $\sum_{j=1}^{n}\sqrt{j}$ and follow the same approach, we end with a multiplicative factor equal to $\color{red}{\frac{9}{4}}$.