Suppose $V$ is finite-dimensional and $<.;.>_1, <.;.>_2$ are inner products on $V$ with corresponding norms $|.|_1$ and $|.|_2$. Prove that there exists a positive number $c$ such that $$|v|_1 \le c*|v|_2$$ for every $v \in V$.
Suppose there exists some $v \not=0 \in V$ such that $|v|_1 > c|v|_2$ for every $c > 0$. However, since this should hold for every positive $c$ then it should hold in particular for $c = |v|_1/|v|_2$. Thus, $|v|_1 > \frac{|v|_1}{|v|_2}|v|_2 = |v|_1$ we get a contradiction.
Is this kind of logic plausible in this case? Note: I saw an answer (If $V$ is a finite dimensional with two norms then $\Vert v\Vert_1 \leq c\Vert v\Vert_2 $) to this question. However, I am trying to avoid using matrices at this point since the book (Linear Algebra Done Right - Axler) is doing so too (at least, I think so). Also, I would like to check if this kind of logic is appropriate.
Here's an approach that doesn't use matrices or anything about compactness.
Suppose that $v_1,\dots,v_n$ is an orthonormal basis relative to $\langle \cdot , \cdot \rangle_2$. Now, consider an arbitrary $v = a_1 v_1 + \cdots + a_n v_n$. We note that $|v|_2 = |a_1|^2 + \cdots + |a_n|^2$. $|\cdot|_1$ is a norm, which means that it satisfies the triangle inequality, so that $$ \begin{align} |v|_1 &= \left|\sum_{j=1}^n a_jv_j\right|_2 \leq \sum_{j=1}^n |a_j v_j|_1 \leq \sum_{j=1}^n |a_j| |v_j|_1 \\ & \leq \sum_{j=1}^n |a_j| \max_{j=1,\dots,n}|v_j|_1 = \max_{j=1,\dots,n}|v_j|_1 \cdot \sum_{j=1}^n |a_j| \end{align} $$ Now, we note that for any non-negative $x_1,\dots,x_n$, we have $$ \frac{x_1 + \cdots + x_n}{n} \leq \sqrt{x_1^2 + \cdots + x_n^2} \implies x_1 + \cdots + x_n \leq n \cdot \sqrt{x_1^2 + \cdots + x_n^2}. $$ Let $M = \max_{j=1,\dots,n}|v_j|_1$. Combining the two inequalities above gives us $$ |v|_1 \leq M(|a_1| + \cdots + |a_n|) \leq M \cdot n \cdot \sqrt{|a_1|^2 + \cdots + |a_n|^2} = M \cdot n \cdot |v|_2. $$ Thus, we have our desired result with $c = M\cdot n$.