How is the Cauchy-Schwarz inequality applied here?

Question

Let $\lambda,\mu$ be measures on a measurable space $(E,\mathcal E)$ and $g:E\to\mathbb R$ be $\mathcal E$-measurable and bounded. It seems to be an application of the Cauchy-Schwarz inequality, but I can't figure out why $$\int\mu({\rm d}x)\int\lambda({\rm d}y)|g(x)-g(y)|^2\ge\frac{\left|\int\mu({\rm d}x)\int\lambda({\rm d}y)\left||g(x)|^2-|g(y)|^2\right|\right|^2}{\int\mu({\rm d}x)\int\lambda({\rm d}y)|g(x)+g(y)|^2}\tag1.$$ So, how do we obtain $(1)$?

Answer 1

First, call $\sigma$ the tensor product of the two measures, so that $$\int f \sigma = \int \lambda(dx) \int \mu(dy) f$$

Call $r(x,y)= |g(x)-g(y)|^2, s(x,y)=|g(x)+g(y)|^2$. Note that $rs(x,y)=|g(x)^2-g(y)^2|^2$. Then you have

$$\left (\int r \sigma \right )\left (\int s \sigma \right ) \ge \int rs \sigma $$

Dividing by the integral in $s$, which is strictly positive, you get the result.