Let $V$ be an inner product space over $F$. Then for all $x,y \in V$ and $c \in F$,

$\|x+y\|$ $\leq$$\|x\| + \|y\|$. (Triangle Inequality)

The following is a proof of the triangle inequality:

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I wanted to know how one gets that $2\Re\langle x,y \rangle \leq 2|\langle x,y\rangle|$, where $\Re $ denotes the real part of the complex number $\langle x,y \rangle$.

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    $\begingroup$ For any complex number $z$, you have that $|z|=\sqrt{Re(z)^2+Im(z)^2}\geq \sqrt{Re(z)^2}=|Re(z)|$, where the inequality is due to removing the non-negative term $Im(z)^2$. $\endgroup$ – user574889 Jul 20 '18 at 2:27

This is nothing to do with inner products, it's just a basic fact about complex numbers, $$|\Re(z)|\le|z|\ .$$ The best way to see this is to draw a diagram, but if you prefer you can do it algebraically: if $z=x+iy$ then $$|z|=\sqrt{x^2+y^2}\ge\sqrt{x^2}=|x|=|\Re(z)|\ .$$


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