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Let $x_i$ be real numbers with $\sum\limits_{i=1}^{n} x_i^2=1$. Prove that $$\frac{1}{2}x_1^2+\frac{1}{6}(x_1+x_2)^2+\cdots +\frac{1}{n(n-1)}(x_1+\cdots +x_{n-1})^2+\frac{1}{n}(x_1+\cdots +x_n)^2\le 3+2\sqrt 2$$

I have tried Abel transformation, induction and other methods with no progress. I also suspect that the left side of this inequality has connection with matrix $a_{ij}=\min (\frac{1}{i}, \frac{1}{j})$, but don't know how to proceed. Thanks for any help.

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    I'm not sure I understand what the general term is, the first one looks like maybe it should be just $x_1^2$ and the second ${1\over 2}(x_1+x_2)^2$ based on what you wrote, or perhaps the general term should be $\displaystyle{1\over (n+1)!}\left(\sum_{i=1}^n x_i\right)^2$ – 2017-01-03
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    Yeah, seems like you want a factorial in the denominator. – 2017-01-03
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    @ThomasAndrews Edited, now it should be clear what the general terms are. :) – 2017-01-03
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    So the last term is an exception? – 2017-01-03
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    @ThomasAndrews yes. – 2017-01-03
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    If $s_k=\sum_{i=1}^{k} x_i$, with $s_0=0$, the left hand side can be written as: $$\sum_{k=1}^{n} \frac{s_{k}^2-s_{k-1}^2}{k}$$ – 2017-01-03
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    That means your connection matrix is right (if I remember what that means) - the left hand side can be written as $\sum_{i,j=1}^{n} \min(1/i,1/j)x_ix_j$. This is because $s_k^2-s_{k-1}^2 = x_k(2x_1+\cdots 2x_{k-1}+x_k)$. – 2017-01-03
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    This can be rewritten as: If $s_1,s_2,\dots$ is a sequence of real numbers which converges, and $\sum_{k=0}^{\infty} (s_k-s_{k-1})^2=S$ for some $S,$ then: $$\sum_{k=0}^{\infty} \frac{s_k^2-s_{k-1}^2}{k}<(3+2\sqrt{2})S$$ – 2017-01-03

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