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Let $a_i>0$, $x_i, y_i\in \mathbb{R}$ $i=1,\cdots, n$, such that

$\sum\limits_{i=1}^nx_iy_i=0$, $\sum\limits_{i=1}^nx_i^2=\sum\limits_{i=1}^ny_i^2=1$. Is it true $$ \left[\sum\limits_{i=1}^n\frac{1}{a_i}x_i^2\right] \left[\sum\limits_{1\le i < j \le n} a_ia_j(x_iy_j-x_jy_i)^2\right]\ge \sum\limits_{i=1}^n a_i y_i^2? $$

I tried the case $n=2$ and the special case where all $a_i$ are equal.

Motivation: To find a lower bound for difference in Lagrange identity under certain conditions.

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@Sunni, congratulations, you get an interesting inequality and a very nice proof. Here are two side observations: 1) in the case $n=2$, it is actually an identity for arbitrary $a_i$'s; 2)when all the $a_i$'s are equal, it is also an identity, for any $n$. –  Syang Chen Jul 25 '11 at 22:04
Right, I already observed that. –  Sunni Jul 26 '11 at 1:42
Note also that from the proof of the inequality you get that equality holds if and only if the vector $(1/\sqrt{a_i}x_i)_{1\leq i\leq n}$ belongs to the linear span of $(\sqrt{a_i}x_i)$ and $(\sqrt{a_i}y_i)$. –  Mikael de la Salle Jul 26 '11 at 7:42
@Mikael, I could see that A is essentially the matrix matrix $\begin{pmatrix}0&\lambda \\ -\lambda&0\end{pmatrix}$ operating on a 2-dimensional subspace $X$ of $R^n$. Thus the identity $\|A\|_{op}=\frac{1}{\sqrt{2}}\|A\|_{HS}$ always holds, and $|Ax'|=\|A\|_{op}|x'|$ if and only if $x'\in X$. But I can't see why $X$ is spaned by $(\sqrt{a_{i}}x_{i})$ and $(\sqrt{a_{i}}y_{i})$. –  Syang Chen Jul 27 '11 at 2:32
@Xianghong, what is immediate is that the kernel of $A$ contains the orthogonal of the space $Y$ spanned by $(\sqrt{a_i} x_i)$ and $(\sqrt{a_i} y_i)$. On the other hand, since $A$ is normal, $X$ is the orthogonal to the kernel of $A$. This proves one inclusion, which is enough since $X$ and $Y$ have same dimension. –  Mikael de la Salle Jul 28 '11 at 5:56

1 Answer 1

up vote 8 down vote accepted

The answer is yes.

Edit: As pointed out in the comment below, my first answer was not correct (it was proving the inequality with a factor 2). Here is the correction.

Denote by $A$ the $n \times n$ matrix given by $A_{i,j} = \sqrt{ a_i a_j} (x_j y_i - x_i y_j)$, by $x'=(\sqrt{a_i^{-1}} x_i) \in \mathbb R^n$ and by $y'=(\sqrt{a_i} y_i) \in \mathbb R^n$. You are asking whether $|y'| \leq |x'| \| A\|_{HS}/{\sqrt 2}$. Here $|\cdot|$ denotes the usual euclidean norm on $\mathbb R^n$, and $\|\cdot\|_{HS}$ is the Hilbert-schmidt norm on matrices, ie $\|M\|_{HS} = \sqrt{Tr(M^*M)}$, where $M^*$ is the (conjugate) transpose of $M$.

By the assumptions $\langle x,y\rangle =0$ and $\langle x,x\rangle =1$, you have that $A x' = y'$. It is therefore enough to prove that $\|A\|_{op} \leq \|A\|_{HS}/\sqrt 2$, where $\|A\|_{op}$ is the operator norm acting on $\mathbb R^n$ with usual euclidean norm. But this holds because $A$ has real entries and is anti-hermitian, ie $A^* = -A$. If $\lambda$ is an eigenvalue of $A$, $\lambda \in i\mathbb R$ and $\bar \lambda$ is also an eigenvalue of $A$. This proves the desired inequality, since $\|A\|_{op}$ is $\max_\lambda |\lambda|$ and $\|A\|_{HS} = (\sum_{\lambda} |\lambda|^2)^{1/2}$, where the max and the sum run over the eigenvalues of $A$ (counted with multiplicity).

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Is your $\otimes$ Kronecker product? I think $\sum\limits_{1\le i < j \le n} a_ia_j(x_iy_j-x_jy_i)^2=\frac{1}{2}\|A\|^2$. –  Sunni Jul 25 '11 at 13:35
You are correct. See my edit. –  Mikael de la Salle Jul 25 '11 at 15:42
+1: nice, clean proof. –  Suvrit Jul 25 '11 at 17:44

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