# Proof: triangle inequality Hermitian inner product

How can I show the triangle inequality for the Hermitian inner product?

I started as follows (using linearity in the second component): \begin{align} 0\leq\lVert v+w\rVert^2&=\langle v+w,v+w\rangle\\ &=\langle v+w,v\rangle +\langle v+w,w\rangle\\ &=\overline{\langle v,v+w\rangle}+\overline{\langle w,v+w}\rangle\\ &=\overline{\langle v,v\rangle}+\overline{\langle v,w\rangle}+\overline{\langle w,v\rangle}+\overline{\langle w,w\rangle}\\ &=\langle v,v\rangle +\overline{\langle v,w\rangle}+\langle v,w\rangle +\langle w,w\rangle\\ &=\lVert v\rVert ++\overline{\langle v,w\rangle}+\langle v,w\rangle+\lVert w\rVert. \end{align} However, what I need is the following: $$\lvert\langle v,w\rangle\rvert\leq\lVert v\rVert \cdot\lVert w\rVert,$$ where $\lvert z\rvert$ stands for the modulus of $z\in\mathbb C$. I feel like I need to go to this form: $$\lvert\langle v,w\rangle\rvert^2=\langle v,w\rangle\cdot\overline{\langle v,w\rangle} \leq\lVert v\rVert \cdot\lVert w\rVert.$$ How can I turn $\overline{\langle v,w\rangle}+\langle v,w\rangle$ into $\langle v,w\rangle\cdot\overline{\langle v,w\rangle}$?

If $\langle v,w \rangle = z$, you have shown that $$\lVert v+w \rVert^2 = \lVert v \rVert^2 + 2\Re{(z)} + \lVert w \rVert^2.$$ Written like this, it is clear that $\lvert z \rvert = \sqrt{\Re(z)^2+\Im(z)^2} \geqslant \Re(z)$, so $$\lVert v \rVert^2 + 2\Re{(z)} + \lVert w \rVert^2 \leqslant \lVert v \rVert^2 + 2\lvert \langle v,w \rangle \rvert + \lVert w \rVert^2.$$ One can then use the Cauchy-Schwarz inequality $\lvert \langle v,w \rangle \rvert \leqslant \lVert v \rVert \lVert w \rVert$ and factorise to finish the job.