Let $f$ be a function defined on the interval $[a,b]$. Expand the inequality of double integral
$\int_a^b\int_a^b[f(x)-f(y)]^2\,dA\geq\,0$
to derive an inequality of 1-variable integration involving $f$
My approach is below：
$\int_a^b\int_a^b[f(x)]^2-2f(x)f(y)+[f(y)]^2\,dx\,dy\geq\,0$
$\int_a^b\int_a^b[f(x)]^2\,dx\,dy-\int_a^b\int_a^b2f(x)f(y)\,dx\,dy+\int_a^b(b-a)[f(y)]^2\,dy\geq\,0$
Let primitive function of $f(x)$ be $F(x)$, I have $\int_a^b\int_a^b2f(x)f(y)\,dx\,dy=\int_a^b2[F(b)-F(a)]f(y)\,dy$
I think $F(b)-F(a)$ is a constant. So the remaining part is $\int_a^b\int_a^b[f(x)]^2\,dx\,dy$
However, I don't know how to change $\int_a^b\int_a^b[f(x)]^2\,dx\,dy$ to 1-variable integration. Do anyone have some ideas? Or my approach is incorrect?
$\int_a^{b}f(x)^{2}dx$ is just a number, not a function of $y$. Hence $\int_a^{b} \int_a^{b}f(x)^{2}dxdy= (\int_a^{b} f(x)^{2}dx)(\int_a^{b} dy)=(b-a)((\int_a^{b} f(x)^{2}dx)$.