We are given $n$ linearly independent vectors $\lbrace \Phi_1, \dots, \Phi_n \rbrace$ with $||{\Phi_i}||=1$. We consider any linear combination $\sum_{i=1}^n \alpha_i \Phi_i$.
My linear algebra script states the following inequality: $\sum \alpha_i^2 - \sum_{i \neq j} |\alpha_i \alpha_j \langle \Phi_i, \Phi_j \rangle| \leq || \sum \alpha_i \Phi_i ||^2$.
I am trying to understand where this inequality comes from, but I don't see it. I think I have to use the inverse triangular inequality, but I am always ending up with a to high term.
\begin{align*} &\sum \alpha_i^2 - \sum_{i \neq j} |\alpha_i \alpha_j \langle \Phi_i, \Phi_j \rangle| \\ \leq &|\sum \alpha_i^2 | - |\sum_{i \neq j} \alpha_i \alpha_j \langle\Phi_i, \Phi_j\rangle| \\ \leq &||\sum \alpha_i^2 | - |\sum_{i \neq j} \alpha_i \alpha_j \langle\Phi_i, \Phi_j\rangle|| \\ \leq &|\sum \alpha_i^2 \langle \Phi_i, \Phi_i \rangle - \sum_{i \neq j} \alpha_i \alpha_j \langle\Phi_i, \Phi_j\rangle| \\ \leq & | \sum_{i,j} \alpha_i \alpha_j \langle \Phi_i, \Phi_j \rangle| \end{align*} But how do I continue without overshooting? Cauchy-Schwarz inequality yields a to large estimation. Or maybe in an earlier step I am already estimating to high?
If you expand $\|\sum \alpha_i\Phi_i\|^{2}$ you get $\sum_{i \neq j} \alpha_i \alpha_j <\Phi_i, \Phi_j>+\sum \alpha_i^{2}$ since $\|\Phi_i\|=1$. Now you only have to observe that $\sum_{i \neq j} |\alpha_i \alpha_j <\Phi_i, \Phi_j>| \geq -\sum_{i \neq j} \alpha_i \alpha_j <\Phi_i, \Phi_j>$ (which is same as $\sum_{i \neq j} \alpha_i \alpha_j <\Phi_i, \Phi_j \geq - \sum_{i \neq j} |\alpha_i \alpha_j <\Phi_i, \Phi_j>|$)