If $V$ is a finite dimensional with two norms then $\Vert v\Vert_1 \leq c\Vert v\Vert_2 $ Suppose $V$ is finite-dimensional and $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ are inner products on $V$ with corresponding norms $\Vert \cdot\Vert_1$ and $\Vert \cdot\Vert_2$. Prove that there exists a positive number $c$ such that
$$\Vert v\Vert_1 \leq c\Vert v\Vert_2 $$
for every $v\in V$.
Please help me with this question?
Edit (to be reopen). Of course, this result is a particular case of the proved result in the linked question. But it is not a duplicate because here we want to use the fact that the norms are induced by inner products. So, is there a different proof for this specific case (where the norms has a specific form)?
 A: Let $(e_1, \dots, e_n)$ be an orthonormal basis with respect to the second scalar product.
Then, for $v = \sum_{i=1}^n a_i e_i$ one has  $\|v \|_2 = \sqrt{\langle v,v \rangle_2} = \sqrt{\sum_{i=1}^n |a_i|^2} \ge \max_{i=1, \dots, n} |a_i| $. 
Now note $\| v\|_1 \le \sum_{i=1}^n |a_i| \| e_i \|_1  \le m \sum_{i=1}^n |a_i| \le mn \max_{i=1, \dots, n} |a_i| \le mn  \|v\|_2$ where $m = \max_{i=1, \dots, n} \|e_i\|_1$. 
A: Let $\{v_1,\dots,v_n\}$ be an orthonormal basis of $V$ with respect to $\langle\, , \, \rangle_1$. One approach is to note that for $x = (x_1,\dots,x_n)\in \Bbb C^n$ and $y = (y_1,\dots,y_n)\in \Bbb C^n$, we have
$$
\left\langle \sum_{i=1}^n x_i v_i, \sum_{j=1}^n y_jv_j \right\rangle_2 = y^*Ax
$$
where $A$ is the matrix whose entries are $a_{ij} = \langle v_i,v_j \rangle$.
From there, it suffices to note (or show) that the function
$$
f(x) = \frac{\|x\|_2}{\|x\|_1} = \frac{x^*Ax}{x^*x}
$$
attains a maximum.  In order to do so, it helps to consider the restriction of $f$ to the set $\{x:\|x\|_1 = 1\}$.
A: Let's assume we're working with a real vector space (proof is similar for a complex vector space). The statement is equivalent to $\langle v,v\rangle_1 \leq c \langle v,v\rangle_2$. 
Every inner product $\langle v,w\rangle_A$ can be written as $v^T A w$ where $A$ is positive definite. Let $A$ be the matrix for inner product 1, and $B$ be the matrix for inner product 2. 
Let the smallest eigenvalue of $B$ be $b$ and the largest eigenvalue of $A$ be $a$. Let $c = \frac{a}{b}>0$. Then,
$$\frac{v^T A v}{v^T v} \leq a,\qquad \frac{v^T B v}{v^T v} \geq b.$$ This follows from the Rayleigh quotient.
So, $$c \frac{v^T B v}{v^T v} \geq c b = a \geq \frac{v^T A v}{v^T v}. $$
Now, multiply both sides by $v^T v$, write $v^T A v = \langle v,v\rangle_1$ and $v^T B v = \langle v,v\rangle_2$ in the expression above, take square roots of both sides and you get your desired statement. 
