Unless otherwise stated, an inner product space will be here either real orthogonal or complex Hermitian. Thus, we can assume that the inner product of a vector, \(v\), with itself, \((v,v)\), is real. In the case that the space \(V\) has a positive definite inner product, then \((v,v)>0\) for any non-zero \(v\) and we can define the length or norm of a vector \(v\) to be
\begin{equation}
\norm{v}=\sqrt{(v,v)}.
\end{equation}
This is a genuine norm, since, as defined, \(\norm{av}=\abs{a}\norm{v}\), \(\norm{v}=0\) implies \(v=0\) and, as we’ll see shortly, \(\norm{v+w}\leq\norm{v}+\norm{w}\).
If, on the other hand, \(V\) has only a non-degnerate inner product, there can exist non-zero vectors, \(v\), with \((v,v)\leq0\). In this case, we could define
\begin{equation}
\norm{v}=\sqrt{|(v,v)|},
\end{equation}
but should note that this is, of course, no longer properly called a norm.

In the positive definite case, we have the important Cauchy-Schwarz inequality.

Theorem (Cauchy-Schwarz inequality) In a positive definite inner product space, \(V\), for any vectors \(v,w\in V\),\begin{equation}
\abs{(v,w)}\leq\norm{v}\norm{w},\label{equ:Cauchy-Schwarz}
\end{equation}
with equality if and only if \(v\) and \(w\) are linearly dependent.

Proof The statement is trivially true when \((v,w)=0\). Assuming \((v,w)\neq0\), so that \(v\neq0\) and \(w\neq0\), for any complex number \(a\) we consider the inner product of the vector \(v-aw\) with itself. Then,\begin{equation*}
(v-aw,v-aw)=\norm{v}^2-a(v,w)-a^*(w,v)+\abs{a}^2\norm{w}^2\geq0,
\end{equation*}
with equality if and only if \(v=aw\), so that choosing
\begin{equation*}
a=\frac{\norm{v}^2}{(v,w)}
\end{equation*}
we have
\begin{equation*}
\norm{v}^2-2\norm{v}^2+\frac{\norm{v}^4\norm{w}^2}{\abs{(v,w)}^2}\geq0
\end{equation*}
so that
\begin{equation*}
\abs{(v,w)}^2\leq\norm{v}^2\norm{w}^2
\end{equation*}
from which the result follows after taking square roots.\(\blacksquare\)

This means that we can define the angle, \(\theta\), \(0\leq\theta\leq\pi/2\) between two vectors, \(v\) and \(w\), in a real orthogonal or complex Hermitian space through,
\begin{equation}
\cos\theta=\frac{\abs{(v,w)}}{{\norm{v}\norm{w}}}.
\end{equation}
The triangle inequality,
\begin{equation}
\norm{v+w}\leq\norm{v}+\norm{w},\label{triangle-standard}
\end{equation}
follows by considering the square of the left hand side and using the Cauchy-Schwarz inequality,
\begin{align*}
\norm{v+w}^2&=\norm{v}^2+\norm{w}^2+2\Real(v,w)\\
&\leq\norm{v}^2+\norm{w}^2+2\abs{(v,w)}\\
&\leq\norm{v}^2+\norm{w}^2+2\norm{v}\norm{w}\\
&=\left(\norm{v}+\norm{w}\right)^2,
\end{align*}
then taking the square root.

Similarly, a couple of variants of the triangle inequality can be obtained by considering the square of the difference of norms,
\begin{align*}
\left(\norm{v}-\norm{w}\right)^2&=\norm{v}^2+\norm{w}^2-2\norm{v}\norm{w}\\
&\leq\norm{v}^2+\norm{w}^2-2\abs{(v,w)}\\
&\leq\norm{v}^2+\norm{w}^2-2\abs{\Real(v,w)}\\
&\leq\norm{v}^2+\norm{w}^2-2\Real(v,w)\\
&=\norm{v-w}^2
\end{align*}
so that taking square roots we obtain
\begin{equation}
\abs{\norm{v}-\norm{w}}\leq\norm{v-w},\label{triangle-variant1}
\end{equation}
or alternatively
\begin{align*}
\left(\norm{v}-\norm{w}\right)^2&=\norm{v}^2+\norm{w}^2-2\norm{v}\norm{w}\\
&\leq\norm{v}^2+\norm{w}^2-2\abs{(v,w)}\\
&\leq\norm{v}^2+\norm{w}^2-2\abs{\Real(v,w)}\\
&\leq\norm{v}^2+\norm{w}^2+2\Real(v,w)\\
&=\norm{v+w}^2
\end{align*}
so that taking square roots we obtain
\begin{equation}
\abs{\norm{v}-\norm{w}}\leq\norm{v+w}.\label{triangle-variant2}
\end{equation}
A further simple consequence of the definition of the norm in terms of the positive definite inner product is the parallelogram identity,
\begin{equation}
\norm{v+w}^2+\norm{v-w}^2=2\left(\norm{v}^2+\norm{w}^2\right),\label{equ:parallelogram}
\end{equation}
which in \(\RR^2\) expresses the fact that the sum of the squared lengths of the diagonals of a parallelogram is equal to twice the sum of the squared lengths of the sides.

We have seen that specifying a positive definite symmetric or Hermitian inner product, \((\cdot,\cdot)\), on a vector space, \(V\), over respectively \(\RR\) or \(\CC\) implies the existence of a norm \(\norm{\cdot}\) on \(V\). That is, real orthogonal and complex Hermitian spaces in which the inner product is positive definite are normed vector spaces. In the other direction, given a normed vector space, \(V\), over \(\RR\), in which the norm, \(\norm{\cdot}\), satisfies the parallelogram identity, \eqref{equ:parallelogram}, we can define a positive definite symmetric inner product as
\begin{equation}
(u,v)=\frac{1}{4}\left(\norm{u+v}^2-\norm{u-v}^2\right).\label{equ:norm inner product}
\end{equation}
Notice that this definition ensures that \(\norm{v}=\sqrt{(v,v)}\). Also, since by the parallelogram identity,
\begin{equation*}
\frac{1}{4}\left(\norm{u+v}^2-\norm{u-v}^2\right)=\frac{1}{2}\left(\norm{u+v}^2-\norm{u}^2-\norm{v}^2\right),
\end{equation*}
and
\begin{equation*}
\frac{1}{4}\left(\norm{u-v}^2-\norm{u+v}^2\right)=\frac{1}{2}\left(\norm{u-v}^2-\norm{u}^2-\norm{v}^2\right),
\end{equation*}
applying the triangle inequality, which the norm satisfies by definition, then leads to the Cauchy-Schwarz identity. To confirm that \eqref{equ:norm inner product} defines a genuine inner product on \(V\) we must check \eqref{inprod-linear}. Using the parallelogram identity we have,
\begin{align*}
\norm{u+v+w}^2&=2(\norm{u+v}^2+\norm{w}^2)-\norm{u+v-w}^2\\
&=2(\norm{u+v}^2+\norm{w}^2)-(2(\norm{u-w}^2+\norm{v}^2)-\norm{u-v-w}^2)\\
&=2\norm{u+v}^2-2\norm{u-w}^2+2(\norm{w}^2-\norm{v}^2)+\norm{u-v-w}^2,
\end{align*}
and
\begin{align*}
\norm{u-v-w}^2&=2(\norm{u-v}^2+\norm{w}^2)-\norm{u-v+w}^2\\
&=2(\norm{u-v}^2+\norm{w}^2)-(2(\norm{u+w}^2+\norm{v}^2)-\norm{u+v+w}^2)\\
&=2\norm{u-v}^2-2\norm{u+w}^2+2(\norm{w}^2-\norm{v}^2)+\norm{u+v+w}^2.
\end{align*}
That is,
\begin{equation*}
\norm{u+v+w}^2-\norm{u-v-w}^2=\norm{u+v}^2-\norm{u-v}^2+\norm{u+w}^2-\norm{u-w}^2
\end{equation*}
so \((u,v+w)=(u,v)+(u,w)\). Next, observe that the definition makes it clear that, \((u,-v)=-(u,v)\), so that,
\((u,nv)=n(u,v)\), for any integer \(n\). It is then easy to see that we must have, \((u,qv)=q(u,v)\), for any \(q\in\QQ\), so that for any \(a\in\RR\) and \(q\in\QQ\),
\begin{align*}
\abs{(u,av)-a(u,v)}&=\abs{(u,(a-q)v)-(a-q)(u,v)}\\
&\leq\abs{(u,(a-q)v)}+\abs{(a-q)}\abs{(u,v)}\\
&\leq2\abs{a-q}\norm{u}\norm{v},
\end{align*}
by Cauchy-Schwartz, and we have that \((u,av)=a(u,v)\) for any \(a\in\RR\).

In the case of a normed vector space, \(V\), over \(\CC\), in which the norm once again satisfies the parallelogram identity we can define a positive definite Hermitian inner product as
\begin{equation}
(u,v)=\frac{1}{4}\left(\norm{u+v}^2-\norm{u-v}^2+i\norm{u+iv}^2-i\norm{u-iv}^2\right).
\end{equation}
The proof that this defines a genuine inner product on \(V\) proceeds as for the real case, the only difference being the real and complex parts are treated separately.