Let $x_1,...,x_n\in X$ a normed vector space and $\|x_i\|=1,\forall i\in\{1,...,n\}$. Suppose that for some $e\in(0,1)$ we have that $\|\sum_{i=1}^n\lambda_ix_i\|\leq (1+e)\max_{i\leq n}|\lambda_i|$ for every real choice of $\lambda_i$. Prove that $\|\sum_{i=1}^n\lambda_ix_i\|\geq (1-e)\max_{i\leq n}|\lambda_i|$.
I can see intuitively that the first relation induces some sort of perpendicularity between the vectors thus restricting the focus on just the largest vector. I cannot see how to proceed though.
The problem can be rephrased as follows: Let $x_i$ be unit vectors s.t. $\|\sum_{i=1}^n\lambda_ix_i\|\leq 1+e$ where $\lambda_i\in [-1,1]$ and at least one $\lambda_i=\pm 1$. Show that $\|\sum_{i=1}^n\lambda_ix_i\|\geq 1-e$ for each choice of $\lambda_i$.
Now suppose wlog that $\lambda_1=1$ otherwise rearrange. Then $2=\|2x_1\|=\|x_1+\sum_{i=2}^n\lambda_ix_i+x_1-\sum_{i=2}^n\lambda_ix_i\|\leq \|x_1+\sum_{i=2}^n\lambda_ix_i\|+\|x_1-\sum_{i=2}^n\lambda_ix_i\|$ and thus if we let $A=\|x_1+\sum_{i=2}^n\lambda_ix_i\|$ and $B=\|x_1-\sum_{i=2}^n\lambda_ix_i\|$ we have $A+B\geq 2$ and thus $B\geq 2-A\geq 2-1-e=1-e$ and the exact same for $A$.