Show that if $F:\mathbb{R}^{n} \to [0,\infty)$ be a Minkowski norm on $\mathbb{R}^{n}$, then $$\lambda^{-1}|(y^{i})|\leq F(y)\leq \lambda |(y^{i})|,$$ where $y:=(y^{i})\in \mathbb{R}^{n}$, $|.|$ is the standard Euclidean norm and $\lambda > 1$ is a constant.
We know two norms $F_{1}$ and $F_{2}$ are said to be equivalent if there are positive numbers $c_{1}$ and $c_{2}$ such that $$c_{1} F_{1}(y)\leq F_{2}(y)\leq c_{2} F_{1}(y) \quad,\quad y\in \mathbb{R}^{n}.$$ Also, a function $F=F(V)$ on a finite dimentional vector space $V$ is called a Minkowski norm if it has the following properties:
(a) $F(y)\geq 0$ for any $y\in V$, and $F(y)=0$ if and only if $y=0$;
(b) $F(ky)= k F(y)$ for any $y\in V$ and $k>0$;
(c) $F$ is $C^{\infty}$ on $V-\{0\}$ such that for any $y\in V$, the following bilinear symmetric functional $g_{y}$ on $V$ is an inner product, $$g_{y}(u,v):= \frac{1}{2}\dfrac{\partial^{2}}{\partial s \partial t}[F^{2}(y+su+tv)]_{s=t=0}$$
Thanks for any help.