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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:

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

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

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

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

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

  6. 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?

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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.

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  • $\begingroup$ Thank you! We can also take $\lambda=\frac{\langle x,y\rangle}{|\langle x,y\rangle|}$ that is the "normalized" inner product. Right? $\endgroup$ – Analysis Newbie Sep 12 '18 at 19:22
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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}

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