Let $S := \{ e_k \}_{k = 1}^{r}$ be an orthonormal set of vectors in $V$, where $r \le \dim(V)$. Let $v \in V$ and let $c_k := \langle v, e_k \rangle$ for $k \in \{1, \ldots, r\}$. Show that $\sum_{k = 1}^{r} c_k^2 \le \| v \|^2$.
Attempt:
Notice that $c_k^2 = (v,e_k)^2 $. We know from Cauchy-Schwarz inequality that $|(v,e_k)| \le ||v|| ||e_k|| = ||v|| $ so that $$ \sum_{k=1}^r c_k^2 \leq \sum_{k=1}^r ||v||^2 = r ||v||^2$$
What am I doing wrong here? or is the inequality not correct??
Let $c_k =\langle v, e_k \rangle$. We have $\|v- \sum\limits_{k=1}^{r}c_k e_k\|^{2}\geq 0$. Expand the left side to get $\|v\|^{2}+ \sum\limits_{k=1}^{r}| \langle v, e_k \rangle|^{2}-2\sum\limits_{k=1}^{r}|\langle v, e_k \rangle|^{2} \geq 0$. Hence $\sum\limits_{k=1}^{r}|\langle v, e_k \rangle|^{2} \leq \|v\|^{2}$