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Let $a_{0},a_{1},\cdots,a_{n}\ge 0,n\ge 1,$ and let $S_{k}=\displaystyle\sum_{i=0}^{k}\binom{k}{i}a_{i}$ with $k=0,1,2,\cdots,n$.

We Assume that $\binom{0}{0}=1,\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$. Prove that

$$\dfrac{1}{n}\displaystyle\sum_{k=0}^{n-1}S^2_{k}-\dfrac{1}{n^2}\left(\displaystyle\sum_{k=0}^{n}S_{k}\right)^2\le\dfrac{4}{45}(S_{n}-S_{0})^2$$

cactus314
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math110
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    is $C_k^i=\dfrac{k!}{(k-i)!i!}$? – Yimin Mar 19 '13 at 07:56
  • yes,Thank you hint – math110 Mar 19 '13 at 08:19
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    Bounties are nice but... what did you try? – Did Jun 02 '13 at 16:15
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    Let's consider the case $a_1 = \dots = a_n = 0$. Then $S_k = a_0$ for all $k$. The inequality reads:

    $$ a_0^2 - \frac{1}{n^2} \big[(n+1)a_0\big]^2 \geq \frac{4}{45} (a_0^2 - a_0^2) = 0 $$

    I suspect you have a typo.

     – cactus314 Jun 09 '13 at 13:39
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    No，$$a^2_{0}-\dfrac{1}{n^2}[(n+1)a_{0}]^2=<0$$ – math110 Jun 09 '13 at 13:54
  • This can be seen as a quadratic form in variables $\vec{a} \in S^n$ whose extreme values are the largest and smallest eigenvalues $\lambda_1 < \vec{a}^T M \vec{a} < \lambda_n$ with $\lambda_1 \leq \dots \leq \lambda_n$. – cactus314 Jun 09 '13 at 15:51

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