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. 



Ask the validity of a scalar inequality 
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).


