Update: The continuity condition is necessary, according to Ian Morris's comment in https://mathoverflow.net/questions/269064/for-a-continuous-function-f-mathbbr-to-mathbbr-does-fx-fy.
Miklós Schweitzer Competition 2015 Problem 8:
Problem 1: Prove that all continuous solutions of the functional equation $$[f(x) - f(y)]\left(f(\tfrac{x+y}{2}) - f(\sqrt{xy})\right) = 0, \forall x, y \in (0, +\infty)$$ are the constant functions.
http://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2015-eng.pdf https://artofproblemsolving.com/community/c6h1224690p6149915 https://mathoverflow.net/questions/269064/for-a-continuous-function-f-mathbbr-to-mathbbr-does-fx-fy
To make the domain and range of $f$ clear (Thank @Calvin Lin for his valuable comments), I rephrased the problem above as follows.
Problem 1 (rephrased): Let $f : \ (0, \infty) \to \mathbb{R}$ be a continuous function satisfying $$[f(x) - f(y)]\left(f(\tfrac{x+y}{2}) - f(\sqrt{xy})\right) = 0, \forall x, y \in (0, +\infty).$$ Prove that $f$ is the constant function.
My question: If the continuity condition on $(0, \infty)$ is necessary?
Any comments are welcome and appreciated.
The solutions in the 2nd link above use the continuity condition. Terry Tao's proof in the 3rd link above also use the continuity condition. My solution (at the end) requires the continuity condition as well. On the other hand, the continuity condition is not necessary for the following problem (from a math exam):
Problem 2: Let $f : (0, \infty) \to (0, \infty)$ be a continuous function satisfying $f(\tfrac{x+y}{2}) - f(\sqrt{xy}) = 0,\ \forall x, y > 0$. Find all $f$.
Solution for Problem 2: $f$ is the constant function. For any $0 < b < a$, let $x = a + \sqrt{a^2 - b^2}$ and $y = a - \sqrt{a^2 - b^2}$. We have $f(a)=f(b)$. [Remark: We do not use the continuity condition in the proof.]
My solution for Problem 1:
Assume, for the sake of contradiction, that $f$ is not the constant function. WLOG, assume there exist two real numbers $0 < A < C$ with $f(A) < f(C)$. Since $f$ is continuous, there exists $A < B \le C$ such that $f(x) < f(B)$ for all $x$ in $[A, B)$.
Consider the sequence $$x_1 = \sqrt{AB}; \ x_{k+1} = \frac{x_k^2 + B^2}{2B}, \ k\ge 1.$$ By using Mathematical Induction, it is easy to prove that $x_k < x_{k+1} < B$ for all $k \ge 1$. Thus, $\lim_{k\to \infty} x_k$ exists. Let $L = \lim_{k\to \infty} x_k$. Then, $L = \frac{L^2 + B^2}{2B}$ and $L = B$. Thus, $\lim_{k\to \infty} x_k = B$.
Denote $a = x_{k+1}, b = x_k$ ($k\ge 1$). Let $X = a - \sqrt{a^2 - b^2}$ and $Y = a + \sqrt{a^2 - b^2}$. It is easy to verify that $X < Y$, $Y = B$, $\frac{X+Y}{2} = a$, and $\sqrt{XY} = b$. Thus, $X = \frac{b^2}{Y} \ge \frac{x_1^2}{B} = A$. Thus, $f(X) - f(Y) \ne 0$. From $[f(X) - f(Y)] [f(\frac{X+Y}{2}) - f(\sqrt{XY})] = 0$, we have $f(a) = f(b)$, i.e., $f(x_k) = f(x_{k+1})$ for all $k \ge 1$. Thus, $f(x_k) = f(x_1) = f(\sqrt{AB})$ for all $k \ge 1$.
Since $f$ is continuous, we have $f(B) = f(\lim_{k\to \infty} x_k) = \lim_{k\to \infty} f(x_k) = f(\sqrt{AB})$. Contradiction.