Proof of Cauchy-Schwarz inequality using a particular result from orthonormal sets

Let {$e_1, ... , e_k$} be an orthonormal set in a finite dimensional real inner product space $V$. If $ v \in V$, I have shown the following result:

$\sum_{i=1}^{k} |\langle v,e_i\rangle|^{2} \leq ||v||^{2} $

I am trying to prove the Cauchy-Schwarz inequality based on this property.

There exists an orthonormal basis {$e_1,...,e_n$} of $V$ and for $v,w \in V$ we therefore have that there exists $a_1,...,a_n,b_1,...,b_n \in \mathbb{R} $ such that $ \sum_{i=1}^{n} a_ie_i = v$ and $ \sum_{i=1}^{n} b_ie_i = w $.

Then $|\langle v,w\rangle| = |\langle v,\sum_{i=1}^{n} b_ie_i\rangle| \\ \qquad \qquad \quad = |\sum_{i=1}^{n} b_i\langle v,e_i\rangle| \\ \qquad \qquad \quad $

I get stuck from here though.

Solution 1:

From you last line, apply the Cauchy-Schwarz inequality: $\mid<v,w>\mid^2\le \displaystyle \sum_{i=1}^kb_i^2\displaystyle \sum_{i=1}^k<v,e_i>^2 \le \displaystyle \sum_{i=1}^kb_i^2||v||^2 = \displaystyle \sum_{i=1}^k <w,e_i>^2||v||^2 \le ||w||^2||v||^2\implies \mid<v,w>\mid\le ||v||||w||.$