Prove that $\left \| \vec{x}-\vec{a} \right \|+\left \| \vec{x}-\vec{b} \right \|=\left \| \vec{a} \right \|-\left \| \vec{b} \right \|$

Question

Let $V$ be a vector space with inner product and $\vec{a},\vec{b}\epsilon V-\left \{ 0 \right \}$.

I'm asked to prove that

if $\left \| \vec{x}-\vec{a} \right \|+\left \| \vec{x}-\vec{b} \right \|=\left \| \vec{a} \right \|-\left \| \vec{b} \right \|$ if and only if, x=na+mb, n+m=1

In the previous question I succesfully proved, using the Parallelogram law, that

$\left \| \vec{a}+\vec{b} \right \|=\left \| \vec{a} \right \|+\left \| \vec{b} \right \|$ if and only if $\vec{a}=λ\vec{b}, λ>0$

Any ideas?

Answer 1

This is not true.

For example, let $V=\mathbb{R}^2,a=x=(1,0)$ and $b=(0,1)$. Clearly $x$ is a linear combination of $a$ and $b$. However, we have $||x-a||+||x-b||=\sqrt2\neq0=||a||-||b||$.