Proving $||f+g||\cdot||f-g|| \le ||f||^2+||g||^2$ in a Hilbert Space

Question

Let $f$ and $g$ be vectors in a Hilbert space $H$.

Show that $$||f+g||\cdot||f-g|| \le ||f||^2+||g||^2$$

My question is, do i have to rewrite $||f+g||$ as $\sqrt{\langle f+g,f+g\rangle}$ and same for $||f-g||$? Or is the product written as $\langle f+g,f-g\rangle?$

I get a very long and undesired result :( any tips will be appreciated. Thank you.

Answer 1

\begin{align} I &= \left \| f + g\right \| \cdot \left \| f-g\right \| = \sqrt{\left \langle f + g, f+g\right \rangle} \cdot \sqrt{\left \langle f - g, f-g\right \rangle} = \\ &= \sqrt{\left \| f\right \|^2 + \left( \left \langle f,g\right \rangle + \left \langle g,f\right \rangle \right )+ \left \| g\right \|^2} \cdot \sqrt{\left \| f\right \|^2 - \left( \left \langle f,g\right \rangle + \left \langle g,f\right \rangle \right ) + \left \| g\right \|^2} \end{align} Assume $\left \langle f,g\right \rangle = a + bi$, then $\left \langle g,f\right \rangle = \overline{\left \langle f,g\right \rangle} = a - bi$, so $\left \langle f,g\right \rangle + \left \langle g,f\right \rangle = 2a$ \begin{align} I &= \sqrt{\left( \left \| f\right \|^2+\left \| g\right \|^2\right )^2 - 4 a^2} \le \sqrt{\left( \left \| f\right \|^2+\left \| g\right \|^2\right )^2} = \left \| f\right \|^2+\left \| g\right \|^2 \end{align}

PS

In case of real Hilbert space $b = 0$, and everything else still holds.