Reverse Triangle Inequality in vector space

Question

I have already proved Triangle Inequality $\lVert v+w\rVert \le \lVert v \rVert+\lVert w \rVert$ in vector space $\mathbb{R}^n$. However, I'm having difficulties proving the Reverse Triangle Inequality in the vector space.

I have started with

$$\lVert v \rVert^2=v \cdot v-w\cdot w+w\cdot w \le \lVert v-w \rVert+\lVert w \rVert^2$$ $$\lVert w \rVert^2=w \cdot w-v\cdot v+v\cdot v \le \lVert w-v \rVert+\lVert v \rVert^2$$

based on the Triangle Inequality but I don't know if it is correct. If this is right, I think I can go on with proving it the same way I would the normal Reverse Triangle Inequality.

Answer 1

$$\lVert v \rVert = \lVert v -w +w \rVert \le \lVert v -w \rVert + \lVert w \rVert$$ $$\lVert w \rVert = \lVert w -v +v \rVert \le \lVert w -v \rVert + \lVert v \rVert = \lVert v-w \rVert + \lVert v \rVert$$

Thus we see that $\lVert v-w \rVert \ge \Big|\lVert v \rVert - \lVert w \rVert\Big|$ as claimed.

Answer 2

Note: This answer was given for what resulted in a duplicated question. Therefore, I removed it from the duplicated question and pasted it here.

In the context of plane geometry, the reverse triangle inequality states that any side of a triangle is greater than or equal to the difference between the other two sides. Namely, there holds that $$ | | u | - | v | | \leqslant | u + v | . $$ The proof follows from the standard triangle inequality (i.e., from the inequality $| u + v | \leqslant | u | + | v |$) as follows. By triangle inequality, you have $$ | u | = | (u + v) - v | \leqslant | u + v | + | v | $$ and, similarly, $$ | v | = | (v + u) - u | \leqslant | u + v | + | u | . $$ But the previous two relations give you that $$ | v | - | u | \leqslant | u + v | $$ and $$ | u | - | v | \leqslant | u + v | . $$ Therefore $| | u | - | v | | \leqslant | u + v |$.