Is this equivalent to Cauchy-Schwarz Inequality?

The Cauchy-Schwarz inequality applies to any semi-definite inner product, including the one on $\mathbb{R}^n$ (as column vectors) defined by $\langle x,y\rangle=x^{\mathrm T}Ay$.

One of the simplest proofs in the real case is to consider $p(t)=\langle x-ty,x-ty\rangle=\langle x,x\rangle -2\langle x,y\rangle t +\langle y,y\rangle t^2$. Then $p(t)\geq 0$ for all $t\in \mathbb R$, and $p$ is a quadratic polynomial in $t$, so its discriminant must be nonpositive. Rearranging this discriminant inequality yields your inequality.

The matrix $A$ has a Cholesky decomposition: $A=L^* L$. Thus

$$x^* A y = (Lx)^* Ly = (Lx) \cdot (Ly),$$ where by $\cdot$ we mean the usual complex inner product: $a \cdot b = \sum_i \overline{a_i} b_i$. Now applying Cauchy-Schwarz,

$$ |x^* A y|^2 \leq ||Lx||_2^2 ||Ly||_2^2 = (x^* L^* L x) (y^* L^* L y) = (x^* A x) (y^* A y)$$