Q:

Inequality involving inner product. $|\langle u,v\rangle+\overline{\langle u,v\rangle}|\le 2|\langle u,v\rangle|$

How is the following inequality involving the inner product true? $|\langle u,v\rangle+\overline{\langle u,v\rangle}|\le 2|\langle u,v\rangle|$
I can see that for a complex number $z=u+iv, z+\overline z=2u$, but is it always true then that $2|u|\le 2|\langle u,v\rangle|$, doesn't this depend on how the inner product is defined? Can this be connected to the triangle inequality?
Thanks in advance!

A:

$$\langle u,v\rangle=x+\mathrm iy\implies|\langle u,v\rangle+\overline{\langle u,v\rangle}|=2|x|=2\sqrt{x^2}\leqslant 2\sqrt{x^2+y^2}=|\langle u,v\rangle|$$

