TITLE: $x \perp y$ if and only if $\Vert x + \alpha y \Vert \ge \Vert x \Vert$ for all scalars $\alpha$
QUESTION [6 upvotes]: Here's Prob. 8 in the Problems after Sec. 3.2 in Introductory Functional Analysis With Applications by Erwine Kreyszig: 

Show that in an inner product space, $x \perp y$ if and only if $\Vert x + \alpha y \Vert \ge \Vert x \Vert$ for all scalars $\alpha$. 

If $x \perp y$, then $\langle x, y \rangle = 0$; so for any scalar $\alpha$, we have 
$$
\begin{align*}
\Vert x + \alpha y \Vert^2 &= \langle x + \alpha y, x + \alpha y \rangle \\
&= \Vert x \Vert^2 + 2 \Re \bar{\alpha} \langle x, y \rangle + \vert \alpha \vert^2 \ \Vert y \Vert^2 \\ 
&= \Vert x \Vert^2 +  \vert \alpha \vert^2 \ \Vert y \Vert^2 \\ 
&\ge \Vert x \Vert^2. 
\end{align*}
$$
So 
$$
\Vert x + \alpha y \Vert \geq \Vert x \Vert.
$$
Am I right?
Now how to prove the converse? It is my guess that we will have to put in a particular value for $\alpha$.

REPLY [5 votes]: $\boldsymbol{\langle x,y\rangle=0\implies\|x+\alpha y\|^2\ge\|x\|^2}$
$$
\begin{align}
\|x+\alpha y\|^2
&=\|x\|^2+2\mathrm{Re}\left(\langle x,\alpha y\rangle\right)+|\alpha|^2\|y\|^2\\
&=\|x\|^2+2\mathrm{Re}\left(\overline{\alpha}\langle x,y\rangle\right)+|\alpha|^2\|y\|^2\\
&=\|x\|^2+|\alpha|^2\|y\|^2\\
&\ge\|x\|^2
\end{align}
$$

$\boldsymbol{\|x+\alpha y\|^2\ge\|x\|^2\implies\langle x,y\rangle=0}$
Let $\alpha=t\langle x,y\rangle$ for $t\in\mathbb{R}$, so that $\langle x,\alpha y\rangle=t|\langle x,y\rangle|^2\in\mathbb{R}$.
Suppose $\langle x,y\rangle\ne0$. For $-\frac2{\|y\|^2}\lt t\lt0$, we have
$$
\begin{align}
\|x+\alpha y\|^2
&=\|x\|^2+2\mathrm{Re}\left(\langle x,\alpha y\rangle\right)+|\alpha|^2\|y\|^2\\
&=\|x\|^2+2t|\langle x,y\rangle|^2+t^2|\langle x,y\rangle|^2\|y\|^2\\
&=\|x\|^2+t|\langle x,y\rangle|^2\left(2+t\|y\|^2\right)\\
&\lt\|x\|^2
\end{align}
$$
Therefore, $\langle x,y\rangle=0$.