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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:

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

with addition not subtraction.

Thank you.

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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 $\|-w\|=\|w\|$.

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