그냥 줏어온 문제 하나.

(2011 SEEMOUS Mathematical Competition, Problem 3) Show that for any three unit vectors \( \hat{a}\), \(\hat{b}\) and \(\hat{c}\), \begin{equation*} \langle \hat{a}, \hat{c} \rangle^2 + \langle \hat{b}, \hat{c} \rangle^2 \leq 1 + | \langle \hat{a}, \hat{b} \rangle| \end{equation*}

Proof. Decompose \begin{align*} \hat{a} &= \cos \alpha \, \hat{c} + \sin \alpha \, \hat{e}_{1}, \hat{b} &= \cos \beta \, \hat{c} + \sin \beta \, \hat{e}_{2}, \end{align*} where \( \hat{e}_1 \) and \( \hat{e}_2 \) are unit vectors orthogonal to \( \hat{c} \). If \( \cos \alpha = 0 \) or \( \cos \beta = 0 \), there is nothing to proof. So assume not and let

\( s = \tan \alpha \), \( t = \tan \beta \) and \( \lambda = \langle \hat{e}_1, \hat{e}_2 \rangle \).

Then the inequality in question reduces to \begin{equation*} \frac{1 - s^2 t^2}{\sqrt{(1 + s^2)(1 + t^2)}} \leq |1 + \lambda s t|. \end{equation*} If \( |st| > 1 \), then this inequality always hold. So it suffices to show this when \( |st| \leq 1 \). In this case, we have \( 1 - |st| \leq | 1 + \lambda s t | \) and it suffices to prove that \begin{equation*} \frac{1 - s^2 t^2}{\sqrt{(1 + s^2)(1 + t^2)}} \leq 1 - |s t|, \end{equation*} which is a direct consequence of Cauchy-Schwart inequality applied to two vectors \( (1, |s|) \) and \( (1, |t|) \).

Corollary. (Generalization of Cauchy-Schwarz Inequality) For any \( a \), \( b \) and \( c \), we have \begin{equation*} | \langle a, c \rangle \langle b, c \rangle| \leq \left( \frac{\| a \| \| b \| + | \langle a, b \rangle |}{2} \right) \| c \|^2. \end{equation*}