# Inequality involving Unit Vectors

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 $$A$$.
• I just tried to play with two vectors and then it occured to me that the second one could be the $\sum_{i=2}^n\lambda_ix_i$. Commented May 12, 2020 at 22:22