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.



A strengthening of the Cauchy-Schwarz inequality
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.


