Inequality relating the probabilities of a quantum state to the euclidean distance of states. My professor has provided us with the following proposition (without proof). ![]( I am trying to prove this. i'm having quite some trouble proving the first inequality, right under the first sentence. Ive tried using the triangle inequality, Cauchy-Schwartz and brute forcing several times but to no avail. I would appreciate any recommendations on how to attempt solving this, while still not providing the entire proof. Note: We use "states" equivalently with "unit vector". To clarify, we are in a finite dimensional Hilbert space (as per Adrian Keister's comment).

Let $\psi^\perp$, $\phi^\perp$ be the projections of $\psi$, $\phi$ on the subspace perpendicular to $x$.

$ |\langle x|\psi\rangle |^2 = 1- \lVert \psi^\perp \rVert ^2 $ and $ |\langle x|\phi\rangle|^2 = 1- \lVert \phi^\perp \rVert ^2 $ so the LHS is $$ \lvert \lVert \psi^\perp \rVert ^2 - \lVert \phi^\perp \rVert ^2 \rvert = \lvert \lVert \psi^\perp \rVert + \lVert \phi^\perp \rVert \rvert \quad \lvert \lVert \psi^\perp \rVert - \lVert \phi^\perp \rVert \rvert \le 2\lvert \lVert \psi^\perp \rVert - \lVert \phi^\perp \rVert \rvert$$ For given lengths, $\lVert \psi^\perp - \phi^\perp \rVert$ is smallest when the vectors are in the same direction, when it equals $\lvert \lVert \psi^\perp \rVert - \lVert \phi^\perp \rVert \rvert$, and $\lVert \psi^\perp - \phi^\perp \rVert \le \lVert \psi - \phi \rVert$