Proving $\sum\limits_{i=1}^k | \langle x,v_i \rangle \langle y,v_i\rangle| \leq \|x\|\cdot \|y\|$ Let $V$ be a real inner product space, and let $v_1,v_2, \dots ,v_k$ be an orthonormal set of vectors. How do you prove that
$$\sum_{i=1}^k | \langle x,v_i \rangle \langle y,v_i\rangle| \leq \|x\|\cdot\|y\|?$$
When does the equality hold?
I've been trying to do this with the Bessel and Cauchy-Schwarz inequalities, but I can't make it work yet. Any help would be greatly appreciated.
 A: We don't necessarily know that $\{v_i:1\le i\le k\}$ is a basis of $V$ since we don't know the dimension of $V$, but we are given that $\{v_i\}$ are orthonormal; that is $\left<v_i,v_j\right>=0$ when $i\not=j$ and $\left<v_i,v_i\right>=1$. 
Consider the vector
$$
x^\perp=x-\sum_{i=1}^k\left<x,v_i\right>v_i\tag{1}
$$
$x^\perp$ is perpendicular to $\{v_i\}$:
$$
\begin{align}
\left<x^\perp,v_j\right>
&=\left<x-\sum_{i=1}^k\left<x,v_i\right>v_i,v_j\right>\\
&=\left<x,v_j\right>-\left<x,v_j\right>\left<v_j,v_j\right>\\
&=0\tag{2}
\end{align}
$$
Therefore, $x^\perp$ is perpendicular to $x-x^\perp=\sum\limits_{i=1}^k\left<x,v_i\right>v_i$.
Next consider
$$
\begin{align}
\left<x-x^\perp,y-y^\perp\right>
&=\left<\sum_{i=1}^k\left<x,v_i\right>v_i,\sum_{j=1}^k\left<y,v_j\right>v_j\right>\\
&=\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\left<v_i,v_i\right>\\
&=\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\tag{3}
\end{align}
$$
Note that since $x^\perp$ is perpendicular to $x-x^\perp$,
$$
\|x-x^\perp\|^2+\|x^\perp\|^2=\|x\|^2\tag{4}
$$
which implies that $\|x-x^\perp\|\le\|x\|$.
Now $(3)$, Cauchy-Schwarz, and $(4)$ yield
$$
\begin{align}
\left|\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\right|
&=\left|\left<x-x^\perp,y-y^\perp\right>\right|\\
&\le\|x-x^\perp\|\|y-y^\perp\|\\
&\le\|x\|\|y\|\tag{5}
\end{align}
$$
To finish off the proof (thanks to cardinal), consider the vector
$$
x^+=\sum_{i=1}^k\left|\left<x,v_i\right>\right|v_i\tag{6}
$$
Note that $\|x^+\|^2=\sum\limits_{i=1}^k\left<x,v_i\right>^2=\|x-x^\perp\|^2$.
Plugging $x^+$ and $y^+$ into $(5)$ gives
$$
\begin{align}
\sum_{i=1}^k\left|\left<x,v_i\right>\left<y,v_i\right>\right|
&=\left|\sum_{i=1}^k\left<x^+,v_i\right>\left<y^+,v_i\right>\right|\\
&\le\|x^+\|\|y^+\|\\
&\le\|x\|\|y\|\tag{7}
\end{align}
$$
