Clarification with a identity

Question

I have the following inequality which comes in proof of triangle inequality.

$\|a+b\|^2=\|a\|^2$+$\|b\|^2+2\Re\langle a|b\rangle\le\|a\|^2+\|b\|^2+2|\langle a|b\rangle |$

I don't know where the first identity comes from and $2\Re\langle a|b\rangle \le 2|\langle a|b\rangle|$ is also unclear to me.

Answer 1

In a complex scalar product, we don't have exact symmetry: $\langle a,b\rangle\ne\langle b,a\rangle$, we only have it conjugated: $$\langle a,b\rangle =\overline{\langle b,a\rangle}\,.$$ So, expanding $(a+b)^2$ instead of $2\langle a,b\rangle$, we get $$\langle a,b\rangle+\langle b,a\rangle=\langle a,b\rangle+\overline{\langle a,b\rangle}=2Re(\langle a,b\rangle)\,.$$ For the other question, any complex number $z=x+yi$ satisfies $|x|=|Re(z)|\le |z|=\sqrt{x^2+y^2}$.

Answer 2

$$||a+b||^2=\langle a+b,a+b\rangle=\langle a,a\rangle + \langle a,b\rangle +\langle b,a\rangle +\langle b,b\rangle=||a||^2+2\operatorname{Re}(\langle a,b\rangle)+||b||^2$$

since $\,\langle b,a\rangle\,=\overline{\langle a,b\rangle}$

Now just check that for $\,z\in\Bbb C\,$ :

$$2\operatorname{Re}(z)\le 2|z|\;\ldots$$