Show that for any three vectors A,B,C in n-space that $||B-A|| \le ||C-A|| + ||C-B||$

Question

If $A,B$ are two vectors in $n$-space, denote by $d(A,B)$ the distance between $A$ and $B$, that is $d(A,B) = ||B-A||$. Show that for any three vectors $A, B, C$ we have $d(A,B) \le d(A,C) + d(B,C)$.

So far I have been able to start by squaring both sides of the inequality, since both sides are positive or zero. Then I have been able to manipulate the inequality down to: $A \cdotp C + C \cdotp B \le C \cdot C + A \cdotp B + ||C-A||*||C-B||$

But I get stuck there. How should I proceed from here, or am I just taking the wrong approach overall?

Answer 1

Since $B-C+C-A=B-A$ and $||-x||=||x||$ it is sufficient to prove that $||x+y||\leqslant ||x||+||y||$.

Now just take $||X+Y||^2=(X+Y;X+Y)=||X||^2+2(X;Y)+||Y||^2$.

While $(||X||+||Y||)^2=||X||+2||X||\cdot||Y||+||Y||^2$

We know that $|(X;Y)|\leqslant||X||\cdot||Y||$ so the conclusion follows

Answer 2

Here are the steps:

• $\vec{AB}=\vec{AC}+\vec{CB}\quad$ Chasles additive relation
• square each side and replace $\vec u\cdot\vec u$ by $\|u\|^2$
• apply Cauchy Schwarz to the $\vec{AC}\cdot\vec{BC}$
• now recognize $(x+y)^2$