An inequality on products of squared norms and dot products

While trying to prove some inequality on the determinant of a matrix, I came up with the following inequality:

Let $u$, $v$, $w$ be three vectors, I would like to show that

$\|u\|^2\|v\|^2\|w\|^2+2\langle u,v\rangle\langle u,w\rangle\langle v,w\rangle \ge \|u\|^2\langle v,w\rangle^2 + \|v\|^2\langle u,w\rangle^2 + \|w\|^2\langle u,v\rangle^2$

I have verified this inequality numerically for vectors of small dimensions using optimization library, so I believe it is correct, but fail to prove so.

It appears to be related to Cauchy-Schwarz inequality, but I don't think it is sufficient for proving it. There is maybe some generalization that I am not aware of. Any help will be appricated.

Solution 1:

Your inequality is juste the non-negativity of the determinant of the Gram matrix of $(u,v,w)$.

The Gram Matrix of a $m$-uple of vectors $(w_1,\ldots,w_m)$ is by definition $\big(\langle w_i,w_j \rangle\big)_{1 \le i,j \le m}$. It si always symmetric and semidefinite positive, so its determinant is non-negative.

Indeed, if $(e_1,\ldots,e_n)$ is any orthonormal basis of any subspace with finite dimension containing $x_1,\ldots,x_m$ and if A denotes the $n \times m$ matrix of $(w_1,\ldots,w_m)$ in the basis $(e_1,\ldots,e_n)$, then the Gram matrix of $(w_1,\ldots,w_m)$ is just the product $A^\top \times A$. And for every $X \in \mathbf{R}^m$ (identified with the space of all real $m \times 1$ matrices), we have $$X^\top \times A^\top \times A \times X = (A \times X)^\top (A \times X) = ||A \times X||_2^2 \ge 0,$$ where $||\cdots||_2$ denotes the canonical euclidian norm on $\mathbf{R}^n$.