Prove that $\Vert\cdot\Vert_2$ and $\Vert\cdot\Vert_3$ are equivalent on $\mathbb R^n$ without using that all norms on $\mathbb R^n$ are equivalent

Question

Prove that $\Vert\cdot\Vert_2$ and $\Vert\cdot\Vert_3$ are equivalent on $\mathbb R^n$ without using the fact that all norms on $\mathbb R^n$ are equivalent.

In other words I need to show that there exist $m,M>0$ such that $$m\Vert x\Vert_3\le\Vert x\Vert_2\le M\Vert x\Vert_3$$ for all $x \in \mathbb R^n$. Where $$\Vert(x_1,\dots,x_n)\Vert_k= \sqrt[k]{\sum_{i=1}^n|x_i|^k}$$

Attempt: I tried using induction to prove that $\Vert x\Vert_2\ge\Vert x\Vert_3$ for all $x\in\mathbb R^n$ but it got really messy, so I am wondering if there is a better way to do this? Also i have no idea how to do the other inequality.

Answer 1

If $\|y\|_2 \leq 1$ then $\|y\|_3 \leq 1$ too because $|y_i| \leq 1$ and so $|y_i|^{3} \leq |y_i|^{2}$. Now take any non-zero $x$ and put $y=\frac x {\|x\|}$ to conclude that $\|x\|_3 \leq \|x\|_2 $. Thus we can take $m=1$.

Next note that $\|y\|_3 \leq 1$ then $\|y\|_2 \leq 1=\sqrt n$ because $|y_i| \leq 1$ for all $i$. Hence we can take $M=\sqrt n$.