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

Question

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!

Answer 1

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

Answer 2

$$|\langle u,v \rangle + \overline{\langle u,v \rangle}| \leq |\langle u,v \rangle| + |\overline{\langle u,v \rangle}| = 2 |\langle u,v \rangle|$$

More generally $|z + \bar z| \leq 2 |z|$ for any $z$.