Let $x, y\in R^n$ and $x, y$ are nonzero, it is well known 
$\frac{x^Ty}{\parallel x\parallel\_2\parallel y\parallel\_2}(\parallel x\parallel\_2+\parallel y\parallel\_2)\le \parallel x+y\parallel\_2$. How to extend this to complex vectors? $arcos\frac{x^Ty}{\parallel x\parallel\_2\parallel y\parallel\_2} $ 
is the angle between $x$ and $y$. What is the appropriate definition of the angle between two complex vectors? I know 
$\frac{\mid x^\*y\mid}{\parallel x\parallel\_2\parallel y\parallel\_2}(\parallel x\parallel\_2+\parallel y\parallel\_2)\le \parallel x+y\parallel\_2$ 
does not hold generally.



What is the angle between two complex vectors?
Let $x,y$ be two nonzero complex vectors, let $\hat x=x/\|x\|$ and $\hat y=y/\|y\|$, and consider the
parabola
$$\phi(t)=\|t\hat x+(1-t)\hat y\|^2=1+2(t^2-t)(1-\Re(\hat x \overline{\hat y})). $$
You easily check that $\phi(t)\ge\phi(1/2)$ for all $t$. This gives the inequality
$$
\|t\hat x+(1-t)\hat y\|\ge \sqrt{\frac{1+\Re(\hat x \overline{\hat y})}2}
$$
for all real $t$. This is stronger than your inequality, which can be obtained by choosing
$$t=\frac{\|x\|}{\|x\|+\|y\|}$$
at the left hand side, and noticing that
$$
\sqrt{\frac{1+\sigma}2}\ge\sigma
$$
for all real $|\sigma|\le1$ at the right hand side. So yes, the correct extension is using $\Re(x \cdot\overline{y})$ instead of $x\cdot y$.


