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.

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 $n=3$.