Assume that $K:\mathbb{R}_{\geq0} \to\mathbb{R}_{\geq0}$ is a decreasing function. If $\nu$ is a probability measure on the unit circle, define its energy $$ \mathcal{E}(\nu):=\int K(|x-y|)d\nu(x)d\nu(y). $$ Is there a simple argument to show that $\mathcal{E}$ is minimized by the (normalized) Lebesgue measure? Assume if necessary that $K$ is bounded.
The obvious idea is to take a measure, rotate it, and say that the linear combination has smaller energy. But this seems to require $K$ to be positive definite, whereas the statement seems to be true more generally.
Update: a potentially useful observation is that any such $K$ can be approximated by positive linear combinations of functions of the form $\mathbb{1}_{x\leq a}$; thus it is sufficient to prove the result for the latter.
Second update: A counterexample has been given in the accepted answer. Still, it would be interesting to see a natural sufficient condition. E. g., what if $K$ is convex (and $|\cdot|$ is the geodesic distance on the circle)?
A counterexample is $K=1_{|x|< 2}$ (where $|\cdot|$ is Euclidean norm) and $\nu=\tfrac12(\delta_1+\delta_{-1}).$ This gives $$\int K(|x-y|)d\nu(x)d\nu(y)=\tfrac12$$ but $$\int K(|x-y|)d\lambda(x)d\lambda(y)=1$$ where $\lambda$ is normalized Lebesgue measure on the unit circle.
Special $K$
$\lambda$ is a minimizer if $K$ is positive semidefinite as a function on the unit circle. There's a proof sketch in the question, or this can be proved by applying Bochner's theorem looking at the Fourier transform.
If $K$ is non-increasing, convex, and $|\cdot|$ is geodesic norm, then $K$ is positive semidefinite, so $\lambda$ is a minimizer. It suffices to consider continuous $K.$ These can be approximated by a piecewise linear non-increasing convex function. These can be written as a non-negative combination of $K(x)=\max(0,c-|x|).$ But $\max(0,c-|x|)$ is positive semidefinite as a function on the unit circle - it's the convolution $1_{x\leq c/2}*1_{x\leq c/2}.$