# Inner Product Spaces – Triangle Inequality

I have to show that:

For an inner product space $V$, $\|x + y \| = \|x\| + \|y\|$, for all $x$, $y \in V$

if and only if one of the vectors $x$ or $y$ is a scalar multiple of the other.

I am thinking, if $x= cy$, for some scalar then the above equality holds.

Edit : I guess have to look at the proof of the triangle inequality which holds as an equality if

$|\langle x, y\rangle | = \|x\|\,\|y\|.$

#### Solutions Collecting From Web of "Inner Product Spaces – Triangle Inequality"

The result is trivial if $x=\mathbf{0}$, as we then have that $x$ is a nonnegative real scalar multiple of $y$ and we have equality. Assume $x\neq \mathbf{0}$.

Look carefully at the proof of the Cauchy-Schwarz inequality:

Express $y$ as $\alpha x + z$, where $\langle x,z\rangle = 0$. This can always be done by letting $\alpha = \frac{\langle x,y\rangle}{\langle x,x\rangle}$ and $z=y-\alpha x$.

Then
\begin{align*} \langle y,y\rangle &= \langle \alpha x+z,\alpha x+z\rangle\\ &= \alpha\overline{\alpha}\langle x,x\rangle + \alpha\langle x,z\rangle + \overline{\alpha}\langle z,x\rangle + \langle z,z\rangle\\ &= |\alpha|^2\lVert x\rVert^2 + \lVert z\rVert^2\\ &\geq |\alpha|^2\lVert x\rVert^2, \end{align*}
with equality if and only if $z=\mathbf{0}$.
Since
$$\alpha = \frac{\langle x,y\rangle}{\langle x,x\rangle} = \frac{\langle x,y\rangle}{\Vert x\rVert^2},$$we conclude that
$$\lVert y\rVert^2 \lVert x\rVert^2 \geq |\langle x,y\rangle|^2$$
with equality if and only if $z=\mathbf{0}$ (i.e., if and only if $y$ is a scalar multiple of $x$).

Now for the triangle inequality, we have
\begin{align*} \lVert x+y\rVert^2 &= \langle x+y,x+y\rangle\\ &= \langle x,x\rangle + \langle y,y\rangle + 2\mathscr{R}e\langle x,y\rangle\\ &\leq \lVert x\rVert^2 + \lVert y\rVert^2 + 2|\langle x,y\rangle| \tag{1}\\ &\leq \lVert x\rVert^2 + \lVert y\rVert^2 + 2\lVert x\rVert\lVert y\rVert \tag{2}\\ &= (\lVert x\rVert + \lVert y\rVert)^2. \end{align*}

Now, $(1)$ is an equality if and only if $|\langle x,y\rangle|=\mathscr{R}e\langle x,y\rangle$, which occurs if and only if $\langle x,y\rangle = \alpha\lVert x\rVert^2$ is a nonnegative real; which occurs if and only if $\alpha$ is a nonnegative real; and $(2)$ is an equality if and only if $z=\mathbf{0}$. Thus, writing $y=\alpha x +z$ with $\langle x,z\rangle = 0$, we conclude that we have equality if and only if $z=\mathbf{0}$ and $\alpha$ is a nonnegative real number.

In summary, equality holds if and only if $x=\mathbf{0}$ (hence is equal to a nonnegative real scalar multiple of $y$), or if $y=\alpha x$ with $\alpha$ a nonnegative real number. That is, if and only if one of $x$ and $y$ is a nonnegative real scalar multiple of the other.

We assume that $x$ and $y$ are not $0$. If we have a relationship of the form $x=cy$ where $c$ is a scalar then $\|x\|=|c|\cdot \|y\|$ so $|c|=\frac{\|x\|}{\|y\|}$. So we try to compute $\Bigl\| \|x\|y-\|y\|x\Bigr\|^2$:
\begin{align*} \Bigl\lVert \|x\|y-\|y\|x\Bigr\rVert^2&=\|x\|^2\|y\|^2+\|y\|^2\|x\|^2-\|x\|\cdot \|y\|(\langle x,y\rangle+\langle y,x\rangle)\\ &=\|x\|\cdot \|y\|(2\|x\|\cdot \|y\|-(\|x+y\|^2-\|x\|^2-\|y\|^2))\\ &=\|x\|\cdot \|y\|(2\|x\|\cdot \|y\|-2\|x\|\cdot \|y\|)\\&=0\end{align*}
which gives the result.