I'm trying to prove Cauchy-Schwarz inequality in an complex inner product space when $\mathbf{x}=\lambda\mathbf y, \lambda\not=0$. but I don't know why the last step means equal:
$$\large|\langle\mathbf \lambda\mathbf y,\mathbf y\rangle|\le \lVert\lambda\mathbf y\rVert\lVert\mathbf y\rVert\\ \large\implies |\lambda||\langle\mathbf y,\mathbf y\rangle|\le|\lambda|\lVert\mathbf y\rVert^2\\ \large\implies|\langle\mathbf y,\mathbf y\rangle|\le\lVert\mathbf y\rVert^2.\ \ \ \ \ \ \ \ $$
Since $\lVert\mathbf y\rVert^2=\langle\mathbf y,\mathbf y\rangle,$ but why it's the same as $|\langle\mathbf y,\mathbf y\rangle|$?
My confusion came from the absolute value of complex number, $|a+bi|=\sqrt{a^2+b^2},$ and I was considering $\langle\mathbf y,\mathbf y\rangle$ be a complex number so I don't know why the equality $\langle\mathbf y,\mathbf y\rangle=|\langle\mathbf y,\mathbf y\rangle|$ will hold.
"Inner product" generally includes "positive definite" as an axiom, so $\langle \mathbf{y},\mathbf{y}\rangle$ is certainly nonnegative, so it equals its absolute value.