Show Cauchy-Schwarz inequality from triangle inequality for $\mathbb{C}$-vector space In many textbooks the proof of Cauchy-Schwarz inequality needs to introduce a parameter $\lambda$ to take $\langle x+\lambda y,x+\lambda y \rangle$ as the first step. I am trying to proof the Cauchy-Schwarz inequality starting from the triangle inequality as follows: 


*

*By triangle inequality of norm: 
$\|x+y\| \le \|x\|+\|y\|$

*Taking square on both sides: 
$\|x+y\|^2 \le (\|x\|+\|y\|)^2$

*The norm induced by inner product: 
$\langle x+y,x+y \rangle \le (\|x\|+\|y\|)^2$

*Expanding both sides: 
$\langle x,x \rangle + \langle x,y \rangle + \langle y,x \rangle + \langle y,y \rangle \le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$

*Eliminating the square norms: 
$\langle x,y \rangle + \langle y,x \rangle \le 2\|x\|\|y\|$

*By the property of inner product: 
$\langle x,y \rangle + \overline{\langle x,y \rangle} \le 2\|x\|\|y\|$
I already have the Cauchy-Schwarz inequality for $\mathbb{R}$-vector space, as $\langle x,y \rangle \equiv \overline{\langle x,y \rangle}$. But for $\mathbb{C}$-vector space, can we finally obtain the Cauchy-Schwarz inequality along this approach? 
 A: Since $\langle x,y\rangle+\overline{\langle x,y\rangle}=2\operatorname{Re}\langle x,y\rangle$, what you proved was that $\operatorname{Re}\langle x,y\rangle\leqslant\lVert x\rVert\lVert y\rVert$. Now, let $\lambda=\dfrac{\bigl\lvert\langle x,y\rangle\bigr\rvert}{\langle x,y\rangle}$ (if $\langle x,y\rangle=0$, that's not a problem, since the inequality is trivial then). So$$\operatorname{Re}\langle\lambda x,y\rangle\leqslant\lVert\lambda x\rVert\lVert y\rVert,\tag1$$but $\operatorname{Re}\langle\lambda x,y\rangle=\operatorname{Re}\bigl\lvert\langle x,y\rangle\bigr\rvert=\bigl\lvert\langle x,y\rangle\bigr\rvert$ and $\lVert\lambda x\rVert=\lVert x\rVert$. Therefore, $(1)$ is the Cauchy-Schwarz inequality.
A: Observe that
\begin{equation}
2Re(\langle x,y\rangle )=\langle x,y\rangle +\overline{\langle x,y\rangle}\leq 2 \Vert x\Vert \Vert y\Vert.
\end{equation}
Now we use the following standard trick. Pick $\alpha$ with $|\alpha|=1$ such that
\begin{equation}
|\langle x,y\rangle|=\alpha \langle x,y \rangle=\langle \alpha x, y\rangle.
\end{equation}
Replacing $x$ by $\alpha x$ in the first inequality yields
\begin{equation}
|\langle x,y\rangle|=Re(|\langle x,y\rangle|)\leq  \Vert \alpha x\Vert \Vert y\Vert=|\alpha| \Vert x\Vert \Vert x\Vert=\Vert x\Vert \Vert y\Vert.
\end{equation}
