It's well known that all norms on a finite dimensional vector space are equivalent, but does restricting to norms arising from inner products make a meaningful difference to how someone could go about proving this result? In other words, can the fact that the norms come from inner products be exploited for an alternative argument?
My motivation for this question comes from Axler's 'Linear Algebra done right' (3rd Ed.). Specifically, question 6.B.12 asks for a proof that two inner products on a finite dimensional vector space produce equivalent norms. Does he intend students to essentially prove the full result, or is there an easier proof for the special case?
Prof. Axler posted a comment which essentially answers this question. Since there are no other answers I will answer my own question based on this comment.
The answer is yes. There is a significantly simpler way to proving the restricted result. The idea is to choose a basis for the vector space that is orthonormal wrt one of the inner products. Given a vector $v$ expressed in terms of the chosen basis, the norm of $v$ wrt the corresponding inner product can just be written down. The result then follows from the triangle inequality, and the fact that if $n$ is a positive integer, and if $x_1,\ldots,x_n\in\mathbb{R}$, then $(x_1+\ldots + x_n)^2\leq n(x_1^2+\ldots + x_n^2)$, which is proven using Cauchy-Schwarz as exercise 6.A.12 in the same book.