Setup: Given a complex inner product space $V$ the Cauchy-Schwarz inequality is $$ |\langle x,y \rangle|\leq \sqrt{\langle x,x \rangle} \sqrt{\langle y,y \rangle}. $$ I know that the RHS is just the product of the induced norms that come from our inner product, but I don't want to pass to any properties of a norm function. I took the standard proof approach (I think?) but I'm not very comfortable with complex variables so I don't know if I'm doing any arithmetic that is invalid over $\mathbb{C}$. Any help would be appreciated!
Proof Attempt: Define a function $f(t) = \langle x+ty,x+ty \rangle$. We can expand this inner product as \begin{align*} \langle x+ty,x+ty \rangle &= \langle x,x+ty\rangle + \langle ty,x+ty \rangle, \textrm{ linear in first arg.} \\ &=\langle x,x\rangle +\langle x,ty \rangle +\langle ty,x \rangle +\langle ty,ty \rangle, \textrm{ linear in second argument.}\\ &=\langle x,x \rangle + \overline{t} \langle x, y \rangle + t\langle y,x \rangle +t\overline{t} \langle y,y\rangle, \textrm{ conjugate linearity in second argument.} \end{align*} Now because $t \in \mathbb{C}$ we know that $t\overline{t} = |t|^2$. We also have by conjugate symmetry that $$ \overline{t}\langle x,y \rangle= t \langle y,x \rangle. $$ Now because the inner product is positive definite, we can conclude that $$ 0 \leq \langle x,x \rangle + 2\overline{t}\langle x,y \rangle +|t|^2 \langle y,y\rangle. $$ Now just like in the case where we are over the reals, I would like to conclude by making the claim that this is a quadratic in $t$ that opens upwards, meaning it has no real roots and hence the discriminant $b^2-4ac \leq 0$ but I don't know if that's valid. It seems fair-ish because $\langle x,x \rangle$ and $\langle y,y \rangle$ are both real numbers, but is there a way to conclude that $\langle x,y \rangle $ is also a real number? I think I need $\langle x,y \rangle$ to be real to apply the quadratic formula right? I suppose intuitively $\langle x,y \rangle $ must be real because there's no ordering on $\mathbb{C}$ so for the inequality to make sense it has to be real?
Maybe more importantly, does it even make sense to think of it as a quadratic because I'm using $|t|^2$ and $\overline{t}$ as the parameters?
The correct formula is $$(x + ty, x + ty) = (x, x) + (ty, ty) + 2\Re((x, ty)) = \|x\|^2 + 2\Re(\bar{t}(x, y)) + |t|^2\|y\|^2.$$ If you want to apply calculus wrt $t$, then let's assume that $t \in \mathbb{R}$. Then we get $$0 \leq (x + ty, x + ty) = \|x\|^2 + 2t\Re((x, y)) + t^2\|y\|^2.$$ Then, since this is a parabola that opens upward and is above the $x$-axis, it has at most one real zero, so $$4\Re((x, y))^2 -4\|y\|^2\|x\|^2\leq 0,$$ i.e. $$|\Re((x, y))| \leq \|x\|\|y\|.$$ Now for any $\alpha \in \mathbb{C}$ with $|\alpha| = 1$, we can replace $x$ with $\alpha x$ to get $$|\Re(\alpha(x, y))| \leq \|x\|\|y\|.$$ Cauchy's inequality comes from $\alpha = \frac{|(x, y)|}{(x, y)}$.