Prove the norm inequality. 
Exercise. Let $p_1$, $p_2$ be norms on $\Bbb R^n$ with respective unit balls $D_!, D_2$. Prove that $$D_2\subset D_1\iff p_1(x)\le p_2(x)\ \forall x\in\Bbb R^n$$

Can someone please help me this. I dont even  know where to start.
 A: Suppose $D_1 \supset D_2$, then $ p_2(x)\leq 1 \implies p_1(x) \leq 1 $
since $p_2\left(\dfrac{x}{p_2(x)}\right) = \dfrac{p_2(x)}{p_2(x)} = 1$, we have $p_1\left(\dfrac{x}{p_2(x)}\right) = \dfrac{p_1(x)}{p_2(x)} \leq 1$
So $p_1(x) \leq p_2(x)$.
The other direction is obvious.
A: Assume $p_1(x)\le p_2(x),\forall x\in\mathbb{R}^n.$ Take a vector $x$ such that $p_2(x)\le 1,$ that is, a vector in the unit ball $D_2.$ Since, $p_1(x)\le p_2(x)$ it is $p_1(x)\le 1.$ So $x\in D_1.$ This shows that $D_2\subset D_1.$
Conversely, assume that $D_2\subset D_1.$ Take $x\in D_2$ such that $p_2(x)=1.$ Since $D_2\subset D_1$ it is $p_1(x)\le 1.$ So, for any $t\ge 0$ it is $p_1(tx)=tp_1(x)\le tp_2(x)=p_2(tx).$ This shows that $p_1(x)\le p_2(x),\forall x\in\mathbb{R}^n.$ 
A: I presume that the balls are closed.
$\Longleftarrow$ 
$x\in D_{2}\Rightarrow p_{2}\left(x\right)\leq1\Rightarrow p_{1}\left(x\right)\leq1\Rightarrow x\in D_{1}$
$\Longrightarrow$ If $D_{2}\subset D_{1}$ then $rD_{2}\subset rD_{1}$
for every $r\geq0$ so that:
$x\in p_{2}\left(x\right)D_{2}\Rightarrow x\in p_{2}\left(x\right)D_{1}\Rightarrow p_{1}\left(x\right)\leq p_{2}\left(x\right)$
Here we took $r=p_2(x)$
