I was working on my Linear algebra homework and I felt I needed to prove this statement (which might not be true), but is at least true for $\mathbb{R}^2$ from a right angled triangle.
Given a inner product space $V$ with derived norm $|| \cdot || = \sqrt{\langle\cdot ,\cdot\rangle}$. Let $a,b,c \in V$. Show that $$ c-a~\bot~b-a \Rightarrow ||c-b|| \geq ||c-a|| $$
Can someone help me proving this?
Using the theorem in our book that states, Given an inner product space $V$ with derived norm $\langle \cdot, \cdot \rangle = \sqrt{||\cdot||}$ we have for $u,v \in V$, if $\langle u,v\rangle = 0$ then $$ ||u + v||^2 = ||u||^2 + ||v||^2 $$ Choosing $u = a-b$ and $v=c-a$ we have $u + v = c - b$ using the theorem we get $$ ||c-b||^2 = ||a-b||^2 + ||c-a||^2 $$ since $||a-b||$ is positive we have $$ ||c-b||^2 \geq ||c-a||^2 \iff ||c-b|| \geq ||c-a|| $$