Normed linear space question Suppose $x_1, x_2, \ldots, x_n$ are linearly independent elements of a normed linear space $X$.  Show that there is a constant $c>0$ with the property that for every choice of scalars $\alpha_1, \ldots, \alpha_n$ we have $$\|\alpha_1x_1+\cdots+\alpha_nx_n\|\geq c(|\alpha_1|+\cdots+|\alpha_n|)$$
I tried doing this by contradiction but I am stuck.
 A: Here's an outline:
1) Use a "normalization argument"  to show that it suffices to prove your result for $|\alpha_1|+\cdots+|\alpha_n|=1$.
2) Define a map from the unit sphere of $\ell_1^n$ to $\Bbb R$ via $(\alpha_1,\ldots,\alpha_n)\mapsto\Vert\alpha_1 x_1+\cdots+\alpha_n x_n \Vert$.
3) Using the axioms of norm, show that this function is continuous.
4) Using the fact that the unit sphere of $\ell_1^n$ is compact, show that this function attains a minimum value.
5) Using the independence of the $x_i$, show that this minimum value is positive.
A: The span $W$ of the vectors $x_k$ is a finite dimensional space.  The mapping
$$\sum_{k=1}^n \alpha_k x_k \mapsto \sum_{k=1}^n |\alpha_k|$$
is a norm. It is well-defined since the vectors $x_k$ form a basis for $W$.   All norms on $W$ are equivalent, so the result follows. 
A: Take the functin $ F: W  \rightarrow \mathbb{R}$ and $W \subset [-1,1] \mathrm{x} [-1,1] \mathrm{x} \cdots [-1,1] \,n$ times $W=\{(\lambda _1,\lambda _2, \cdots , \lambda _n) |  |\lambda _1 |+ |\lambda _2 |+ \cdots |\lambda _n |=1 \}$ $W$ is closed hence compact times such that $F(y)= \|\lambda_1x_1+\cdots+\lambda_nx_n\|$ now$F$ is continuous $F(y) >0$ and  for every choise of $y \in W$ and since is compact it takes a minimum that is your $c$
A: Try to solve the problem in case where $X$ is an inner product space.
Notice that we can assume without loss of generality that $x_i$ span $X$, and that all finite-dimensional normed spaces look a lot like inner product spaces (since all norms on a finite-dimensional space are equivalent).
In fact, you can probably choose an inner product where all the vectors in question are orthogonal.
A: If there exist scalars $\alpha_1, \ldots, \alpha_n$ not all zero such that for all $c > 0$ we have
$$\left\|\sum_{i=1}^n\alpha_i x_i\right\| < c\sum_{i=1}^n\lvert \alpha_i\rvert,$$
then $\alpha_1 x_1 + \cdots + \alpha_n x_n = 0$, which means the $x_i$ are not linearly independent.
