Summary of my question: Does $\| \mathbf a \wedge \mathbf b\| \leq \| \mathbf a \| \| \mathbf b\|$ hold for all $\mathbf a \in \Lambda^k(\mathbb R^n)$ and $\mathbf b \in \Lambda^\ell(\mathbb R^n)$?
Some background: Given $\mathbb R^n$ with the standard inner product, we can define an inner product on the exterior powers $\Lambda^k(\mathbb R^n)$ by $\langle u_1 \wedge \cdots \wedge u_k, v_1 \wedge \cdots \wedge v_k\rangle = \det(\langle u_i, v_j\rangle)_{i,j=1}^k$ and then extending by linearity. As usual, for $\mathbf a \in \Lambda^k(\mathbb R^n)$, let $\|\mathbf a\| = \sqrt{\langle \mathbf a, \mathbf a \rangle}$. If $\mathbf a = u_1 \wedge \cdots \wedge u_k$ (i.e., if $\mathbf a$ is simple), then $\|\mathbf a \|$ is equal to the volume of the $k$-dimensional parallelotope generated by $u_1, \ldots, u_k$.
When $\mathbf a$ and $\mathbf b$ are simple, then we can see that $\| \mathbf a \wedge \mathbf b\| \leq \| \mathbf a \| \| \mathbf b\|$ holds by using the geometric interpretation given above.
When $\mathbf a = u_1 \wedge \cdots \wedge u_k + v_1 \wedge \cdots \wedge v_k$ and $\mathbf b$ is simple, then after some lengthy computations (expanding the inner product and Gram-Schmidt), I was able to show the inequality holds in this case as well.
Is the inequality $\| \mathbf a \wedge \mathbf b\| \leq \| \mathbf a \| \| \mathbf b\|$ true in general? And is there a geometric interpretation of these quantities when $\mathbf a$ and $\mathbf b$ are not simple?
I had a similar question, and found a general bound here (p. 9, equation 1.10) given by $$ \|\mathbf{a} \wedge \mathbf{b}\| \le \begin{pmatrix} k + \ell \\ k \end{pmatrix}^{1/2} \|\mathbf{a}\|\|\mathbf{b}\|. $$ The proof of this fact can be found here (Geometric Measure Theory, Federer: Section 1.7.5).