# Triangle Inequality lower and upper bounds?

I was solving a question for linear algebra today and had a question regarding triangle inequalities. The question is:

If $$\Vert \vec{v} \Vert = 5$$ and $$\Vert \vec{w} \Vert = 3$$, what are the smallest and largest values of $$\Vert \vec{v} - \vec{w} \Vert$$?

The way that I solved it is to draw two circles with the origin as their centers and each having radius $$5$$ and $$3$$. In this case, $$\vec{v}$$ and $$\vec{w}$$ would be each circle's radius. It's not hard to see that the smallest value we can obtain is $$2$$ and the largest is $$8$$.

However, I attempted to try solving this question using the triangle inequality. I noticed that the way to solve it is $$| \Vert \vec{v} \Vert - \Vert \vec{w} \Vert | \le \Vert \vec{v} - \vec{w} \Vert \le \Vert \vec{v} \Vert + \Vert \vec{w} \Vert$$.

The first part I recognize as being the reverse triangle inequality, but how was the second part derived? In my head the original form of the inequality is:

$$\Vert \vec{v} + \vec{w} \Vert \le \Vert \vec{v} \Vert + \Vert \vec{w} \Vert$$

The minimum is $$2$$ and maximum is $$8$$. Hint: take $$v =cw$$ where $$c$$ is a scalar to see that these values are actually possible.
$$\|v-w\|=\|v+(-w)\|\leq \|v\|+\|w\|$$ because $$\|-w\|=\|w\|$$.