Cauchy-Schwarz inequality for nonsymmetric matrices I am interested in Cauchy-Schwarz inequalities for inner products of the form $\langle Ax,y\rangle$ where $A$ is some matrix. It is rather easy to see that the Cauchy-Schwarz inequality continues to hold when relaxing the assumption of positive definiteness on $A$ to positive semi-definitess. I asked myself whether one can loosen the assumption on symmetry of $A$.
Suppose $A \in \mathbb{R}^{n \times n}$ is a matrix such that $\langle Ax,x\rangle \ge \lambda |x|^2$. Is it true that for all $x,y \in \mathbb{R}^n$ we have the following inequality
$$\langle Ax,y \rangle^2 \le \langle Ax, x \rangle \langle Ay,y \rangle? $$
If not, is there any other similar inequality, which replaces the Cauchy-Schwarz inequality?
 A: No, not even when $\lambda>0$. Consider $A=\lambda I+K$ where $\lambda$ is a small positive number and $K$ is a nonzero skew-symmetric matrix. Then $\langle Ax,x\rangle=\lambda\|x\|^2$ for all vectors $x$. However, when $x$ is chosen such that $y=Kx\ne0$, we have
\begin{aligned}
\langle Ax,y\rangle^2
&=\left(\lambda\langle x,y\rangle+ \langle Kx,y\rangle\right)^2\\
&=\left(\lambda\langle x,y\rangle+\|y\|^2\right)^2\\
&>\lambda^2\|x\|^2\|y\|^2\\
&=\langle Ax,x\rangle\langle Ay,y\rangle
\end{aligned}
when $\lambda$ is sufficiently small.
The inequality in question is true when $A$ is symmetric and $\lambda\ge0$. In this case the condition $\langle Ax,x\rangle\ge\lambda\|x\|^2$ implies that $A$ is positive semidefinite. Therefore $(x,y)\mapsto\langle Ax,y\rangle$ defines a semi-inner product and Cauchy-Schwarz inequality is satisfied.
Alternatively, since the quadratic form $\langle Ax,x\rangle$ depends only on the symmetric part $S=\frac{A+A^T}{2}$ of $A$, when there is some $\lambda\ge0$ such that $\langle Ax,x\rangle\ge\lambda\|x\|^2$ for all $x$, we see that $S$ is positive semidefinite. Therefore, Cauchy-Schwarz inequality gives $\langle Sx,y\rangle^2\le\langle Sx,x\rangle\langle Sy,y\rangle$ and consequently,
$$
\left(\frac{\langle Ax,y\rangle+\langle x,Ay\rangle}{2}\right)^2
\le\langle Ax,x\rangle\langle Ay,y\rangle.
$$
