Prove that there does not exists $w\in V$ such that $||v-w||\le4$ and $||u-w||\le4$.

Question

Suppose let $u,v\in V$ , $||u||=2$ and $||v||=11$ .Prove that there does not exists $w\in V$ such that $||v-w||\le4$ and $||u-w||\le4$.

how to prove this statement i think we prove by contradiction

i.e suppose there exists a $w\in V$ such that $||v-w||\le4$ and $||u-w||\le4$.

i can't prove this contradiction .

Answer 1

Hint: use some triangle inequality.

If $u$ is distance $2$ from $0$ and $w$ is distance at most $4$ from $u$, then what is the maximum distance $w$ can be from $0$?

If $v$ is distance $11$ from $0$ and $w$ is distance at most $4$ from $v$, then what's the minimum distance $w$ can be from $0$?

Answer 2

think about the meaning of $\vec \ell_1=\vec u-\vec w$. this means that $\vec\ell$ is the vector that connect $\vec u$ and $\vec w$. therefore given $\|\vec u\|=2$ and that $\sup\|\vec\ell_1\|=4$ we can calculate $\sup\|\vec w\|=6$. now let's look at it from $\vec v$. let's $\vec\ell_2=\vec v-\vec w.$ we can calculate $\inf\|\vec\ell_2\|$: if $\vec v=t\vec w$. now we know that $\|\vec v\|=11$ then $\inf\|\vec\ell_2\|=11-\sup\|\vec w\|=11-6=5$. and because $\inf\|\vec\ell_2\|=5>4$ we conclude that it is impossible.