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Suppose $\mathbf{v},\mathbf{w} \in \mathbb{R}^n$ (and if it helps, you can assume they each have non-negative entries), and let $\mathbf{v}^2,\mathbf{w}^2$ denote the vectors whose entries are the squares of the entries of $\mathbf{v}$ and $\mathbf{w}$.

My question is how to prove that \begin{align*} \|\mathbf{v}^2\|\|\mathbf{w}^2\| - \langle \mathbf{v}^2,\mathbf{w}^2\rangle \leq \|\mathbf{v}\|^2\|\mathbf{w}\|^2 - \langle \mathbf{v},\mathbf{w}\rangle^2. \end{align*}

Some notes are in order:

  • The Cauchy-Schwarz inequality tells us that both sides of this inequality are non-negative. Thus the proposed inequality is a strengthening of Cauchy-Schwarz that gives a non-zero bound on the RHS.
  • I know that this inequality is true, but my method of proving it is extremely long and roundabout. It seems like it should have a straightforward-ish proof, or should follow from another well-known inequality, and that's what I'm looking for.
Nathaniel Johnston
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    Consider Newton's inequalities for symmetric functions. Gerhard "This Seems Related To That" Paseman, 2018.06.02. – Gerhard Paseman Jun 02 '18 at 17:35
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    This seems to be highly dependent on the symmetric structure of the $L^2$-space $\mathbb{R}^n = \ell^2({1,\dots,n})$. In fact, the inequality is not in general true on the weighted $L^2$-space $\ell^2({1,\dots,n}, \mu)$. Counterexample: for $n=2$, $\mu = (1, 1/2)$, $v = (1,2)$ and $w = (1,1)$ we obtain $LHS = 3\sqrt{3/2} - 3 \approx 0.67$ and $RHS = 1/2$. – Jochen Glueck Jun 02 '18 at 20:50
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    About 2 or 3 years ago I saw an strength version of Cauchy Schwarz inequality in researchgate introduced by George Stoica. I was connected to him via RG but i can not find him neither in RG nor in the internet. I would like to add his observation to this post, as a comment and I wish to inform him of this interesting MO post but I can not find him.Any way +1 for your very interesting post. – Ali Taghavi Jun 02 '18 at 22:21
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    @AliTaghavi Do you mean this article: A Cauchy-Schwarz type inequality inspired by Statistics? – polfosol Jun 03 '18 at 08:57
  • I think that I'm being silly, but in what sense is this a strengthening of CS? I assume you mean that CS can be deduced from it (more easily than just by proving CS directly), but I don't see how. – LSpice Jun 03 '18 at 20:30
  • @LSpice - I just mean a strengthening of CS in the sense of the first bullet point above. CS is equivalent to the statement that RHS $\geq 0$. However, this inequality is stronger since it replaces the $0$ lower bound by something that is typically bigger than $0$. – Nathaniel Johnston Jun 03 '18 at 23:07
  • Somehow I totally missed that bullet point; sorry. I'm still epistomelogically troubled, since the non-negativity of the LHS uses CS, but that's just a terminological, not a mathematical, issue. – LSpice Jun 03 '18 at 23:40
  • @polfosol Yes yes exactly But I loosed that file and I do not remember his result. Where is he now? – Ali Taghavi Jun 04 '18 at 10:10
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    For cross-referencing's sake, this result was used (and this MO thread cited) in our recent paper https://arxiv.org/abs/1807.06897 – Nathaniel Johnston Jul 19 '18 at 14:17

2 Answers2

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Here is a proof for every $n$. Using the notation $\mathbf{v}=(v_1,\dots,v_n)$ and $\mathbf{w}=(w_1,\dots,w_n)$, the inequality reads $$\left(\sum_i v_i^4\right)^{1/2}\left(\sum_i w_i^4\right)^{1/2}-\sum_i v_i^2 w_i^2\leq \left(\sum_i v_i^2\right)\left(\sum_i w_i^2\right)-\left(\sum_i v_i w_i\right)^2.$$ Rewriting the right hand side in a familiar way, and then rearranging and squaring, we obtain the equivalent form $$\left(\sum_i v_i^4\right)\left(\sum_i w_i^4\right)\leq\left(\sum_i v_i^2 w_i^2+\sum_{i<j}(v_iw_j-v_j w_i)^2\right)^2.$$ Rewriting the left hand side in a familiar way, we obtain the equivalent form $$\left(\sum_i v_i^2w_i^2\right)^2+\sum_{i<j}(v_i^2w_j^2-v_j^2w_i^2)^2\leq\left(\sum_i v_i^2 w_i^2+\sum_{i<j}(v_iw_j-v_j w_i)^2\right)^2.$$ Equivalently, $$\sum_{i<j}(v_i^2w_j^2-v_j^2w_i^2)^2\leq 2\left(\sum_k v_k^2w_k^2\right)\sum_{i<j}(v_iw_j-v_j w_i)^2+\left(\sum_{i<j}(v_iw_j-v_j w_i)^2\right)^2.$$ It will be clear in a moment why we renamed the variable $i$ to $k$ in the first sum on the right hand side. Namely, we claim that the following stronger inequality holds: $$\sum_{i<j}(v_i^2w_j^2-v_j^2w_i^2)^2\leq 2\sum_{i<j}(v_i^2w_i^2+v_j^2w_j^2)(v_iw_j-v_j w_i)^2+\sum_{i<j}(v_iw_j-v_j w_i)^4.$$ Indeed, this inequality can be rearranged to $$0\leq 2\sum_{i<j}(v_iw_i-v_jw_j)^2(v_iw_j-v_jw_i)^2,$$ and we are done.

GH from MO
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In the case $n=2$ this follows from the identity below, which expresses the difference

$$\bigl(||\mathbf{v}||^2 ||\mathbf{w}||^2 - \langle \mathbf{v}, \mathbf{w} \rangle^2 + \langle \mathbf{v}^2,\mathbf{w}^2\rangle\bigr)^2 - \bigl(||\mathbf{v}^2|| || \mathbf{w}^2|| \bigr)^2$$

as a product of two squares

$$ \bigl( (v_1^2+v_2^2)(w_1^2+w_2^2) - (v_1w_1+v_2w_2)^2 + (v_1^2w_1^2+v_2^2w_2^2)^2 \bigr)^2 - (v_1^4+v_2^4)(w_1^4+w_2^4) = 2(v_1w_2-v_2w_1)^2(v_1w_1-v_2w_2)^2. $$

For the Cauchy–Schwarz inequality the analogous argument generalizes to any dimension. However I have not been able to extend this argument even to $n=3$.

Mark Wildon
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