Deriving the identity $\langle\phi_1|\phi_1 \rangle + \langle\phi_2|\phi_2 \rangle \geq 2Re(\langle\phi_1|\phi_2 \rangle)$

Question

Consider two arbitrary vectors $|\phi_1 \rangle$ and $| \phi_2 \rangle$ belonging to the inner product space:

My notes then state the following inequality holds but I have no idea where it came from:

$$\langle\phi_1|\phi_1 \rangle + \langle\phi_2|\phi_2 \rangle \geq 2Re(\langle\phi_1|\phi_2 \rangle)$$

Any tips on how to understand where this come from would be appreciated.

Answer 1

Take two vectors $v$ and $w$ which belong to the underlying inner product space.

Thus we have that:

\begin{align*} \langle v - w,v -w\rangle & = \langle v,v\rangle - \langle v,w\rangle - \langle w,v\rangle + \langle w,w\rangle \geq 0\\\\ & \Rightarrow \langle v,v\rangle + \langle w,w\rangle \geq \langle v,w\rangle + \overline{\langle v,w\rangle} = 2\text{Re}\langle v,w\rangle \end{align*}

and we are done.

Hopefully this helps!

Answer 2

$$\Re\langle\phi_1|\phi_2 \rangle\,\leq\,|\langle\phi_1|\phi_2 \rangle|\,\leq\,||\phi_1||\,||\phi_2||\,\leq\,\frac{||\phi_1||^2+||\phi_2||^2}{2}.$$

The first step is a property of complex numbers. The second is Cauchy-Schwarz. Lastly, for any two reals, their product is bounded above by the average of their squares.