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 $x^2+y^2\ge 2xy,x+2y\ge 2y$
 A: 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 $\color{red}{\frac{9}{4}}$.
