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Does any one knows the correct version of the inequality: Let $a_1,\cdots,a_n\in \mathbb R$ and $b_1,\cdots,b_n\in \mathbb R$ with $\sum_{j=1}^n b_j=0$. Then $$ \left(\sum_{j=1}^n a_jb_j \right)^2\leq \left(\sum_{j=1}^n a_j^2-\left(\sum_{j=1}^n a_j\right)^2\right)\sum_{j=1}^n b_j^2\,, $$

proposed in https://link.springer.com/content/pdf/10.1007%2F978-1-4939-1926-0_2.pdf Exercise 2.51, page 23?

Indeed, the statement as such is incorrect. e.g. taking $n=2$ and $a_1=a_2=1$, the statement yields $$ 0\leq 2-2^2. $$

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  • $\begingroup$ Your counterexample needs to include $b_j$ as well. $\endgroup$
    – eyeballfrog
    Jul 25, 2022 at 14:50
  • $\begingroup$ @eyeballfrog Any non trivial $b_j$ will work as the l.h.s vanishes, e.g. $b_1=1$ and $b_2=-1$ $\endgroup$
    – Tok Tak
    Jul 25, 2022 at 14:54
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    $\begingroup$ Upon further reflection, that inequality is always false if all the $a$ are positive. Since $\sum_{j=1}^n a_j^2 - (\sum_{j=1}^n a_j)^2 =-2 \sum_{i\ne j}a_ia_j$, the RHS is negative while the LHS is nonnegative. Very curious. That said, that difference of squares usually occurs in the context of variances, where it would be $n^{-1}\sum_{j=1}^n a_j^2 - (n^{-1}\sum_{j=1}^n a_j)^2$, a quantity that is always positive. Perhaps that's where the problem arises? $\endgroup$
    – eyeballfrog
    Jul 25, 2022 at 14:59

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I believe the issue is that rather than sums, those should be means. That is, $$ \left[\frac{1}{n}\sum_{j=1}^na_j^2 - \left(\frac{1}{n}\sum_{j=1}^na_j\right)^2\right]\left(\frac{1}{n}\sum_{j=1}^n b_j^2\right)\ge\left(\frac{1}{n}\sum_{j=1}^na_jb_j\right)^2 $$ This can also be written more compactly in vector notation: $$ \left[\frac{|\mathbf{a}|^2}{n}-\left(\frac{\mathbf{a}\cdot\mathbf{1}}{n}\right)^2\right]\frac{|\mathbf{b}|^2}{n}\ge \left(\frac{\mathbf{a}\cdot\mathbf{b}}{n}\right)^2, $$ where $\mathbf{1}$ is the vector of all $1$'s We now use the well-known statistics identity $\langle (x - \langle x\rangle)^2\rangle = \langle x^2\rangle - \langle x\rangle ^2$ to deduce that $|\mathbf{a}|^2/n - (\mathbf{a}\cdot \mathbf{1}/n)^2 = |\mathbf{a}-(\mathbf{a}\cdot\mathbf{1}/n)\mathbf{1}|^2/n$. Then we apply Cauchy-Schwartz to the LHS: \begin{multline} \left[\frac{|\mathbf{a}|^2}{n}-\left(\frac{\mathbf{a}\cdot\mathbf{1}}{n}\right)^2\right]\frac{|\mathbf{b}|^2}{n} = \frac{1}{n^2}\left|\mathbf{a}-\left(\frac{\mathbf{a}\cdot\mathbf{1}}{n}\right)\mathbf{1}\right|^2|\mathbf{b}|^2\\\ge\frac{1}{n^2}\left[\mathbf{a}\cdot\mathbf{b}-\frac{(\mathbf{a}\cdot\mathbf{1})(\mathbf{1}\cdot\mathbf{b})}{n}\right]^2 = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{n}\right)^2, \end{multline} where the last equality comes from $\sum_{j=1}^n b_j = 0$, i.e., $\mathbf{1}\cdot\mathbf{b} = 0$.

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  • $\begingroup$ Thanks @eyeballfrog, this is indeed a correct possible "correction"! I wonder whether there might be a version of the statement which may be extented to (infinite) sequence $\endgroup$
    – Tok Tak
    Jul 26, 2022 at 9:50

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