Showing that $\sqrt{\langle \mathbf{a}, \mathbf{a} \rangle}$ satisfies triangle inequality.

Question

The following is a proof from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:

With regards to the proof for property 4 (triangle inequality), I'm confused about the following:

1. In the last section of the proof, the author goes from $\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}$ to $2||\mathbf{a}|| ||\mathbf{b}||$. I'm assuming the author is using the preceding derivation of $|\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}| \le 2||\mathbf{a}|| \cdot ||\mathbf{b}||$. However, I don't understand how this substitution is valid, since, in this derivation, the author uses $|\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}|$, which is the absolute value, instead of just $\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}$, as would be required. In fact, I'm not sure why the author uses the expression with absolute values instead of without, since we originally had the one without -- the absolute values were just thrown in with no justification.

2. For the triangle inequality, we need to show that $||\mathbf{a} + \mathbf{b}|| \le ||\mathbf{a}|| + ||\mathbf{b}||$. But in the last section of the proof, I cannot see anything that resembles this. All we have is $||\mathbf{a} + \mathbf{b}||^2 \le ||\mathbf{a}||^2 + 2||\mathbf{a}||||\mathbf{b}|| + ||\mathbf{b}||^2$. I just don't see how this is proving the triangle inequality?

I would greatly appreciate it if people could please take the time to clarify these two points.

Answer 1

$(1)$: As the author noted, $\langle a, b\rangle + \overline{\langle a, b\rangle}$ is real. For every real number $r$ we have $r\leqslant |r|$.

$(2)$: You missed the power of two in the LHS. It's ${\lVert a + b\rVert}^2$.

Answer 2

1. You are partly right, but we can use that $\alpha:=\langle a, b\rangle+\langle b, a\rangle$ as any real number satisfies $\alpha\le|\alpha|$.
   Note also that we want to use the Cauchy-Schwartz inequality $|\langle a, b\rangle|\le\|a\|\cdot\|b\|$, and that justifies introducing the absolute value sign.
2. It's $\|a+b\|^{\bf 2}$ which is less or equal than the right hand side, which amounts to $(\|a\|+\|b\|)^2$.