Functional equation $f\left(\frac{x+y}{2}\right)+f\left(\frac{2xy}{x+y}\right)=f(x)+f(y)$ implies $2f(\sqrt{xy})=f(x)+f(y)$

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Prove that if the function $f$ is defined on the set of positive real numbers, its values are real, and $f$ satisfies the equation $$f\left( \frac{x+y}{2}\right) + f\left(\frac{2xy}{x+y} \right) =f(x)+f(y)$$ for all positive $x,y$, then $$2f(\sqrt{xy})=f(x)+f(y)$$ for every pair $x,y$ of positive numbers.


Source: Miklos Schweitzer Memorial Competition 2001


I can see how the repeated application of the functional equation condition upon itself forms a bound, but how can I formally prove this?