This is my problem:
For any choice of a couple of orthonormal vectors $\alpha,\beta\in \mathbb{R}^4$ and for any $u,v\in \mathbb{R}$, I have to prove the inequality $v^2(\alpha_1^2+\beta_1^2-1)+u^2(\alpha_2^2+\beta_2^2-1)-2uv(\alpha_1\alpha_2+\beta_1\beta_2)\le 0$
In my notation $\alpha=(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ and $\beta=(\beta_1,\beta_2,\beta_3,\beta_4)$, so we know $\sum_{j=1}^4\alpha_j^2=\sum_{j=1}^4\beta_j^2=1$ and $\sum_{j=1}^4\alpha_j\beta_j=0$.
As you may notice the components $\alpha_3,\alpha_4,\beta_3,\beta_4$ play a subtle role in the inequality. I wasn't able to prove it in an effective way for any choice of $\alpha,\beta,u,v$ so I was hoping someone would help me.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$Hints: Because $(\alpha, \beta)$ is an orthonormal pair, for every vector $U$ in $\Reals^{4}$, the orthogonal projection of $U$ to the plane spanned by $(\alpha, \beta)$ is $$ U' := \Brak{U, \alpha} \alpha + \Brak{U, \beta} \beta, $$ and $$ \|U'\|^{2} = \Brak{U, \alpha}^{2} + \Brak{U, \beta}^{2} \leq \|U\|^{2}. $$ Take $$ U = (v, -u, 0, 0), $$ and note that your inequality is equivalent to $$ (v\alpha_{1} - u\alpha_{2})^{2} + (v\beta_{1} - u\beta_{2})^{2} \leq u^{2} + v^{2}. $$