Let $X$ be Hilbert space. Prove the following statements are equivalent:
1) $x \perp y, $ $x,y \in X$
2) $||x||\le||x+ty||$ for every $t\in \mathbb{C}$
3) $||x-ty||=||x+ty||$
Obtaining 2) from 1) was fairly easy. However I can't find proper way using inner product to get 3) from 2).
First note that $$\|x\pm ty\|^2 = \|x\|^2 \pm 2\operatorname{Re}(t\langle y,x\rangle) + |t|^2\|y\|^2\tag{*}$$
$\underline{1)\implies 2)}$
From $(*)$ we get $\|x+ty\|^2 = \|x\|^2 + |t|^2\|y\|^2\geq \|x\|^2$ for all $t\in\Bbb C$.
$\underline{2)\implies 1)}$
From $(*)$ and $(2)$ we get $2\operatorname{Re}(t\langle y,x\rangle) + |t|^2\|y\|^2\geq 0$ for all $t\in\Bbb C$. Let $t = \lambda\langle x,y\rangle$ for some $\lambda\in\Bbb C$.
Assuming $\langle x,y\rangle \neq 0$, we have
\begin{align} 2\operatorname{Re}(t\langle y,x\rangle) + |t|^2\|y\|^2\geq 0 &\implies 2\operatorname{Re}(\lambda|\langle x,y\rangle|^2) + |\lambda|^2|\langle x,y\rangle|^2\|y\|^2\geq 0\\ &\implies 2\operatorname{Re}\lambda + |\lambda|^2\|y\|^2\geq 0,\ \forall\lambda\in\Bbb C \end{align}
For real $\lambda$ this means that $$\lambda(\lambda\|y\|^2+2)\geq 0.$$ Now choose any $\lambda \in (-\frac 2{\|y\|^2}, 0)$ to arrive at contradiction.
$\underline{1)\iff 3)}$
From $(*)$ we have that $\|x-ty\| = \|x+ty\|$ if and only if $\operatorname{Re}(t\langle y, x\rangle) = 0$. Thus, $1)$ obviously implies $3)$.
Now assume that $\operatorname{Re}(t\langle y, x\rangle) = 0$, for all $t\in\Bbb C$. For $t = 1$, we get $\operatorname{Re}(\langle y, x\rangle) = 0$, while for $t = -i$ we get $\operatorname{Im}(\langle y, x\rangle) = \operatorname{Re}(-i\langle y, x\rangle) = 0$. Thus, $\langle x,y\rangle = 0$.