Prove that for any vectors $\vec{u}$ and $\vec{v}$ of an euclidean vector space and any $\lambda, \mu \in \mathbb{R}$, you have that $$\left\|\lambda\vec{u}+\mu\vec{v}\right\|\leq|\lambda|\cdot\left\|\vec{u}\right\|+|\mu|\cdot\left\|\vec{v}\right\|$$
I'm not sure if I did it correct? We have
$$\left\| \lambda \vec{u} + \mu \vec{v}\right\| \leq |\lambda| \cdot \sqrt{\left\langle \vec{u}, \vec{u} \right\rangle} + |\mu| \cdot \sqrt{\left\langle \vec{v}, \vec{v} \right\rangle} \Leftrightarrow$$
$$\left\| \lambda \vec{u} + \mu \vec{v}\right\| \leq \sqrt{|\lambda|^2 \cdot \left\langle \vec{u}, \vec{u} \right\rangle} + \sqrt{|\mu|^2 \cdot \left\langle \vec{v}, \vec{v} \right\rangle} \Leftrightarrow$$
$$\left\| \lambda \vec{u} + \mu \vec{v}\right\| \leq \sqrt{\lambda \cdot \bar{\lambda} \cdot \left\langle \vec{u}, \vec{u} \right\rangle} + \sqrt{\mu \cdot \bar{\mu} \cdot \left\langle \vec{v}, \vec{v} \right\rangle} \Leftrightarrow$$
$$\left\| \lambda \vec{u} + \mu \vec{v}\right\| \leq \sqrt{\lambda \cdot \left\langle \vec{u}, \lambda \vec{u} \right\rangle} + \sqrt{\mu \cdot \left\langle \vec{v}, \mu \vec{v} \right\rangle} \Leftrightarrow$$
$$\left\| \lambda \vec{u} + \mu \vec{v}\right\| \leq \sqrt{\left\langle \lambda \vec{u}, \lambda \vec{u} \right\rangle} + \sqrt{\left\langle \mu \vec{v}, \mu \vec{v} \right\rangle} \Leftrightarrow$$
$$\left\| \lambda \vec{u} + \mu \vec{v}\right\| \leq \left\| \lambda \vec{u} \right\| + \left\| \mu \vec{v} \right\|$$
And now at the last step I'm not sure what to do? Is it possible to form it further than that? Or there is another, easier way of doing this? :s
That looks good so far, and just recall that the triangle inequality for norms: $\|a + b \| \leq \|a\| + \|b\|$, so the final inequality is always true.