Suppose $1 \leq k \leq n$ and $x_1, x_2, \ldots, x_k$ are $k$ vectors in $\Bbb R^n$ satisfying for any $1 \leq i, j \leq k$, $\langle x_i, x_j\rangle = \delta_{ij}$. For each $1 \leq j \leq k$, let $a_j$ be the 1st component of $x_j$ . Show that $\sum_{j=1}^k a^2_j\geq 1$. I am kind of getting a feel that I have to use least square method but unable to figure it out. Please help.
Edit: The above inequality will be reversed as Kavi Rama Murty observed. I actually made a typo but don't want to change the question drastically as suggested by the math people. So I am adding this that we have to prove $\sum_{j=1}^k a^2_j\geq 1$.
This is false. Take $n=3, x_1=(0,1,0), x_2=(0,0,1)$.
EDIT: The reverse inequality follows by Parseval's identity Extend $\{x_1,x_2,...,x_k\}$ to an orthonomal basis $\{x_1,x_2,...,x_n\}$ and conclude that $1=\|e_1\|^{2}=\sum\limits_{i=1}^{n} |\langle e_1, x_i \rangle|^{2} \geq \sum\limits_{i=1}^{k} |\langle e_1, x_i \rangle|^{2}$ where $e_1=(1,0,0,,,0)$.