Is this equivalent to Cauchy-Schwarz Inequality? $(x^*Ay)^2 \leq (x^*Ax)(y^*Ay)$
Does this inequality hold for positive semi-definite matrix A? I wonder if this is equivalent to Cauchy-Schwarz Inequality. I tried to diagonalize A but it still half-way towards Cauchy-Schwarz Inequality.
Any thoughts on this inequality? Am I going on the right direction?
Thanks a lot!
 A: 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.
A: 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, 
$$ |x^* A y|^2 \leq ||Lx||_2^2 ||Ly||_2^2  = (x^* L^* L  x) (y^* L^* L y) = (x^* A x) (y^* A y)$$
A: Yes, it is equivalent to CS inequality. W.l.o.g, asumme $A$ is a diagonal matrix with positive diagonal entries, $A=diag(a_1,\cdots, a_n)$. you will see that $\left(\sum \sqrt{a_i}x_i\sqrt{a_i}y_i\right)\le \left(\sum (\sqrt{a_i}x_i)^2\right)\left(\sum (\sqrt{a_i}y_i)^2\right) $  is in the same form of CS inequality.
