$\lVert tx +y\rVert\ge 0$, for all $t\in\mathbb R$, implies that $\,|\langle x,y\rangle|\le \lVert x\rVert \lVert y \rVert$

Question

Let $t \in \mathbb{R}$ and $x,y \in \mathbb{C^n}$ and $0 \leq \lVert tx +y\rVert$. How can I transform $\lVert tx +y\rVert$ in something like $at^2 + bt +c$ ?

And then I want to look at the discriminant $b^2 -4ac$, from there I can show $0 \leq \lVert tx +y\rVert \Rightarrow |\langle x,y \rangle| \leq \lVert x\rVert \lVert y\rVert$.

For $x,y \in \mathbb{R^n}$ I have proof it by myself, but for $x,y \in \mathbb{C^n}$ I can't.

Answer 1

Consider $$ 0\le\|tx+y\|^2=(tx+y,tx+y)=\|x\|^2t^2+2(x,y)t+\|y\|^2=at^2+bt+c. $$ Since, $at^2+bt+c\ge 0$, for all $t\in\mathbb R$, then the roots of $at^2+bt+c=0$ are either complex of real and equal, and hence its discriminant $\Delta=b^2-4ac$ is non-positive, i.e. $$ 0\ge b^2-4ac=4(x,y)^2-4\|x\|^2\|y\|^2, $$ which implies that $$ |(x,y)|\le \|x\|\|y\|. $$