Let $(X,\|\cdot\|)$ be a (complex) normed vector space, and let $Y$ be a closed subspace of $X$. Given an element $x_0\in X$ which is not in $Y$, since $Y$ is closed, the distance between $x_0$ and $Y$ is positive. Let $\delta$ denote this positive distance. We thus have $ \|y-x_0\|\geq \delta$ for all $y\in Y$. Consider the points in $\operatorname{span}(Y\cup\{x_0\})$. They can be uniquely written in the form of $y+tx_0$, where $y\in Y$ and $t\in\mathbf{C}$. Then what can we say about their norm?
A simple observation gives $$ \|y+tx_0\|=|t|\|y/t+x_0\|\geq \delta|t|.$$ It seems that we should also have $$ \|y+tx_0\|\geq C\|y\|$$ for some constant $C>0$ (not depending on $y$ and $t$). I tried to prove it, but I didn't succeed. I'd like to know whether the claim is true or not. Can someone give some hints？
If $\|y\|\geq 2t \|x_0\|$, then the claim follows by the triangle inequality. So assume $\|y\| < 2t \|x_0\|$. Suppose that for all $n\in\mathbb{N}_{\geq 2}$, there exist $y_n\in Y$ and $t_n\in \mathbb{C}$ such that $$\frac{\|y_n+t_nx_0\|}{\|y_n\|}<\frac{1}{n},$$ then $$\|y_n+t_nx_0\|<\frac{2t_n}{n}\|x_0\|,$$ which implies that $x_0\in Closure(Y)$.