Show that : $\|x\|^{2}\ge\sum\limits_{i=1}^{n}|\langle x,x_{i}\rangle |^2$ Let $\mathbb{H}$ be a Hilbert space and $x_{1},x_{2},x_{3},...,x_{n}$ orthogonal vectors from $\mathbb{H}$ , $x\in\mathbb{H}$ then prove that : 

$$\|x\|^{2}≥\displaystyle\sum_{i=1}^{n}|\langle x,x_{i}\rangle |^{2}$$

For all $x\in\mathbb{H}$
I don't know how I start  in the proof, I don't have any hints in this type of questions.
If someone know a book or PDF where I can find like this problem?
 A: Hint: Write $$x=x_0 +\sum_{j=1}^n \langle x, x_j \rangle x_j$$ where $x_0$ is in the orthogonal complement of $\operatorname{span}\{ x_1, \dots, x_n \}$. Then use $\Vert x \Vert^2 =\langle x, x \rangle $ and use that $x_j\perp x_i$ for $i,j\in \{0, \dots , n\}$ and $i\neq j$.
A: We wish to show that $$\frac{\sum_{i=1}^n|\langle x,x_i\rangle|^2}{\|x\|^2}\leq 1$$
$$LHS=\frac{\sum_{i=1}^nx^Tx_ix_i^Tx}{\|x\|^2}$$
$$=\frac{x^T(\sum_{i=1}^nx_ix_i^T)x}{\|x\|^2}$$
$$=\frac{x^TAx}{x^Tx}$$
where $A$ is an idempotent matrix.
EDIT 1
We know that maximum value of $\frac{x^TAx}{x^Tx}$ is absolute value of the maximum eigen value of $A$, which is $1$ here, because $A$ is an idempotent matrix.
EDIT 2
I am assuming that the vectors $x_i,1\leq i\leq n$ are orthonormal, the proposition is not true otherwise, as noted in comments by Peter Melech and Severin Schraven. Also, the above argument is valid only for finite dimensional Hilbert Space.
A: Noting
$$ 0\le \|x-\sum_{i=1}^n\left<x,x_i\right>x_i\|^2=\|x\|^2-2\left<x,\sum_{i=1}^n\left<x,x_i\right>x_i\right>+\|\sum_{i=1}^n\left<x,x_i\right>x_i\|^2 $$
one has
$$ 2\sum_{i=1}^n\left<x,x_i\right>^2\le\|x\|^2+\|\sum_{i=1}^n\left<x,x_i\right>x_i\|^2. $$
Since
$$ \|\sum_{i=1}^n\left<x,x_i\right>x_i\|^2=\sum_{i=1}^n\left<x,x_i\right>^2 $$
one has
$$ \sum_{i=1}^n\left<x,x_i\right>^2\le\|x\|^2. $$
