Use the law of cosines to derive the triangle inequality

Question

I am given the vectors:

and show that they span the triangle with sides $a,b,c$ with $c=||u-v||$

and determine for which $\gamma∈[0, \pi]$ we have equality.

Any help is appreciated.

Answer 1

Note that the length of the sides of the triangle are $\|u\| = a$, $\|v\| = b$ and $\|u - v\| = c$. Also, the cosine of the angle between the vectors $u$ and $v$ is $\frac{u \cdot v}{\|u\|\|v\|} = \frac{ab\cos(\gamma)}{ab} = \cos(\gamma)$ which we can easily confirm graphically since $u$ is just along the horizontal axis. Now, the cosine law says $c^2 = a^2 + b^2 - 2ab\cos(\gamma)$. Since $\cos(\gamma) \geq -1 \implies -\cos(\gamma) \leq 1$, so $$ c^2 = a^2 + b^2 - 2ab\cos(\gamma) \leq a^2 + b^2 + 2ab = (a + b)^2. $$ Since both $c$ and $a + b$ are positive we can take square roots and still preserve the inequality. So $c \leq a + b$ or $\|u - v\| \leq \|u\| + \|v\|$. Looking at the above inequality, we see that its an equality precisely when $\cos(\gamma) = -1$ which implies $\gamma = \pi$.

Answer 2

I am assuming that you are looking to show that $c^{2} \leq a^{2} + b^{2}$. If not I apologise, but this could still be helpful

Let $c^{2} = || u - v ||^{2} \\ = (u-v, u-v) \\ = (u, u-v) - (v, u-v) \\= (u,u) - (u,v) + (v,v) - (v,u) \\ = ||u||^{2} + ||v||^{2} - 2(u,v). $

Now $||u||^{2} =a^{2}$ and $||v||^{2} = b^{2} (\cos^{2}\gamma + \sin^{2}\gamma) = b^{2}$. Note that $(u,v) = ||u||\cdot||v||\cos\gamma$ where $\gamma$ is the angle between $u$ and $v$. Note, for $\gamma \in [0,\pi]$, that $\cos\gamma \in [0,1]$, which is positive. Hence we can conclude that

$c^{2} = a^{2} + b^{2} - 2||u||\cdot||v||\cos\gamma \leq a^{2} + b^{2}, \quad \gamma \in [0,\pi] $

and equality is gained when $\cos\gamma = 0$