# Prove $l^2$ norm obeys the triangle inequality

I'm trying to work through Exercise 3 from this blog post, which is essentially a proof of the validity of the $$l^2$$ norm:

Exercise 3: Let $$(\mathcal{V},\left<\cdot,\cdot\right>)$$ be an inner product space. Show, that $$||x|| = \sqrt{\left}\ \forall x \in \mathcal{V}$$ is a normed vector space.

So far my working is as follows (I apologise for the hand-written working, I can convert it to MathJax if required by the site standards);

Which is where I get stuck. Can anyone advise how to finish showing the triangle inequality holds for the $$l^2$$ norm? I think I may need to use the Schwartz inequality, but I'm not sure how to apply it in the context.

• Note that the LHS in your last step is $$\langle x,y \rangle + \langle y,x \rangle = \langle x,y \rangle + \overline{\langle x,y \rangle} = 2 \Re \langle x,y \rangle,$$ and conclude using the Schwartz inequality. (Recall also that $\Re z \leq |z|$ for any complex number.) – MisterRiemann Feb 24 at 12:20
• Aha! I was closer than I realised! Thanks @MisterRiemann! – aaronsnoswell Feb 25 at 6:20

Any inner product satisfies the C-S inequality $$|\langle x, y \rangle| \leq \|x\| \|y\|$$ from which you can complete your argument immediately.