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

Question

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 $x^2+y^2\ge 2xy,x+2y\ge 2y$

Answer 1

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

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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 $k\ge1/2$ $$ (k+1/2)^{3/2}-(1/2)^{3/2}\gt k^{2/3}(k+1/2)^{5/6}\tag{2} $$

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

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

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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, $$ \begin{align} \sum_{k=1}^n\raise{4pt}{\frac{k^2}{\sum\limits_{j=1}^ka_j^2}} &\le\sum_{k=1}^n\frac1{\sigma(k)}\left(\sum_{j=1}^k\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &=\sup_{\|c\|_2=1}\left(\sum_{k=1}^n\frac{c_k}{\sigma_k^{1/2}}\sum_{j=1}^k\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &=\sup_{\|c\|_2=1}\left(\sum_{j=1}^n\sum_{k=j}^n\frac{c_k}{\sigma_k^{1/2}}\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &\le\left(\sum_{j=1}^n\left(\sum_{k=j}^n\frac1{\sigma_k}\right)^{\!\!1/2}\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &\le\left(\sum_{j=1}^n\frac32j^{-3/4}\frac{j^{3/4}}{a_j}\right)^{\!\!2}\\ &=\frac94\left(\sum_{j=1}^n\frac1{a_j}\right)^{\!\!2}\tag{6} \end{align} $$

Answer 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 $\color{red}{\frac{9}{2}}$.

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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 $\color{red}{\frac{9}{4}}$.

Answer 3

Although I believe that the optimal constant is $1$. I was only able to prove the inequality up to a multiplicative factor of $\color{blue}{\frac{3}{2}}$,

$$ \sum_{k=1}^{n} \frac{k^2}{a_1^2 + a_2^2 + \cdots + a_k^2} \leq \color{blue}{\frac{3}{2}} \Biggl( \sum_{k=1}^{n} \frac{1}{a_k} \Biggr)^2, $$

a slight improvement from other answers.

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Step 0 - Integral Analogue. Its integral analog is much easier to prove. For positive increasing function $f$,

\begin{align*} \left( \int_{0}^{\infty} \frac{1}{f(x)} \, \mathrm{d}x \right)^2 &= \int_{0}^{\infty}\frac{2}{f(r)} \biggl( \int_{0}^{r} \frac{1}{f(t)} \, \mathrm{d}t \biggr) \mathrm{d}y \\ &\geq \int_{0}^{\infty} \frac{2r}{f(r)^2} \, \mathrm{d}r \\ &= \int_{0}^{\infty} \frac{6r^4}{f(r)^2} \left( \int_{r}^{\infty} \frac{\mathrm{d}x}{x^4} \right) \, \mathrm{d}r \\ &= \int_{0}^{\infty} \frac{6}{x^4} \left( \int_{0}^{x} \frac{r^4}{f(r)^2} \, \mathrm{d}r \right) \mathrm{d}x \\ &\geq \int_{0}^{\infty} \frac{6}{x^4} \frac{\left( \int_{0}^{x} r^2 \, \mathrm{d}r \right)^2}{\int_{0}^{x} f(r)^2 \, \mathrm{d}r} \mathrm{d}x \tag{$\because$ C–S} \\ &= \frac{2}{3} \int_{0}^{\infty} \frac{x^2}{\int_{0}^{x} f(r)^2 \, \mathrm{d}r} \mathrm{d}x. \end{align*}

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Step 1. First, we extend the sequence by letting $a_{n+k} = +\infty$ for each $k \geq 1$. This allows us to convert the finite sums to infinite series, thereby lessening some notational load. Next, we rearrange $(a_k)_{k=1}^{\infty}$ in increasing order to obtain an increasing sequence $(b_k)_{k\in\mathbb{N}}$. Then we rearrange the sum as

\begin{align*} \left( \sum_{k=1}^{n} \frac{1}{a_k} \right)^2 &= \left( \sum_{k=1}^{\infty} \frac{1}{b_k} \right)^2 = \sum_{k,l=1}^{\infty} \frac{1}{b_k b_l} = \sum_{r=1}^{\infty} \sum_{\substack{j, k \geq 1 \\ \max\{j,k\}=r}} \frac{1}{b_k b_l} . \end{align*}

Step 2. Now for each $r \geq 1$, there are precisely $2r-1$ pairs $(j, k)$ for which $\max\{j,k\} = r$ holds, and for each of such pairs, we have $b_k b_l \leq b_r^2$. So

\begin{align*} \left( \sum_{k=1}^{n} \frac{1}{a_k} \right)^2 \geq \sum_{r=1}^{\infty} \frac{2r-1}{b_r^2} &= \sum_{r=1}^{\infty} \frac{(2r-1)r^3}{b_r^2} \sum_{j=r}^{\infty} \left( \frac{1}{j^3} - \frac{1}{(j+1)^3} \right) \tag{1} \\ &= \sum_{j=1}^{\infty} \left( \frac{1}{j^3} - \frac{1}{(j+1)^3} \right) \sum_{r=1}^{j} \frac{(2r-1)r^3}{b_r^2}. \end{align*}

Step 3. We now get into the dirty part of analyzing the sum in the last step. By the Cauchy–Schwarz inequality,

$$ \Biggl(\sum_{r=1}^{j} \frac{(2r-1)r^3}{b_r^2} \Biggr)\Biggl( \sum_{r=1}^{j} b_r^2 \Biggr) \geq \Biggl( \sum_{r=1}^{j} \sqrt{(2r-1)r^3} \Biggr)^2. \tag{2} $$

Plugging this back and noting that $ \sum_{r=1}^{j} a_r^2 \geq \sum_{r=1}^{j} b_r^2 $, we are led to

$$ \left( \sum_{k=1}^{n} \frac{1}{a_k} \right)^2 \geq \sum_{j=1}^{\infty} \frac{j^2} {\sum_{r=1}^{j} a_r^2} \cdot \underbrace{ \frac{1}{j^2} \left( \frac{1}{j^3} - \frac{1}{(j+1)^3} \right) \Biggl( \sum_{r=1}^{j} \sqrt{(2r-1)r^3} \Biggr)^2 }_{=(*)} $$

Step 4. So it suffices to prove that $(*)$ is bounded below by $\frac{2}{3}$. By noting that $f(x) = \sqrt{2 - x}$ is concave on $[0, 1]$, we get $f(x) \geq x f(1) + (1-x) f(0) $. Then plugging $x = 1/r$ gives

$$ \sqrt{(2r-1)r^3} \geq \sqrt{2} r^2 - (\sqrt{2} - 1) r $$

for $r \geq 1$. Summing both sides for $r = 1, 2, \ldots, j$,

$$ \sum_{j=1}^{r} \sqrt{(2r-1)r^3} \geq \sum_{r=1}^{j} \left( \sqrt{2} r^2 - (\sqrt{2} - 1) r \right) = \frac{1}{6} j (j + 1) \bigl( 2\sqrt{2} j + (\sqrt{2}-1)^2 \bigr). $$

So, squaring both sides, multiplying $\frac{1}{j^2}\Bigl( \frac{1}{j^3} - \frac{1}{(j+1)^3} \Bigr)$, and substituting $x = 1/j \in [0, 1]$ gives

\begin{align*} (*) &\quad \geq \frac{1}{36} \biggl( \frac{3j^2 + 3j + 1}{j^3(j+1)} \biggr) \bigl( 2\sqrt{2} j + (\sqrt{2}-1)^2 \bigr)^2 \\ &\quad = \frac{1}{36} \biggl( 3 + \frac{x^2}{1+x} \biggr) \bigl( 2\sqrt{2} + (\sqrt{2}-1)^2 x \bigr)^2 \end{align*}

This is increasing in $x$, hence is further bounded below by

$$ \frac{1}{36} \cdot 3 \cdot \bigl( 2\sqrt{2} \bigr)^2 = \frac{2}{3}. $$

Therefore the claim follows.

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Remarks.

1. Here is my thought as to why we pick up an extra multiplicative factor.

   Note that $(1)$ is rather conservative, becoming precise when $b_1 = b_2 = \cdots = b_n$. On the other hand, the Cauchy–Schwarz inequality, $(2)$ will become closer to an equality if $b_r \asymp r$. So, the steps $(1)$ and $(2)$ are optimized for different types of sequences, thereby rendering the entire argument unoptimized.

2. In $\text{(1)}$, we utilize the equality

   $$ 1 = r^p \sum_{j=r}^{\infty} \left( \frac{1}{j^p} - \frac{1}{(j+1)^p} \right) $$

   with $p = 3$. Although one may play with the values of $p$, the choice $p = 3$ optimizes the resulting multiplicative constant.