Assume the contrary that for all $a, b \in \mathbb{R}^+$ we have that $f(a) + f(b) \le 2f(\sqrt{ab})$. Let $P(a, b)$ denote that assertion. First, $P(a, \frac{1}{a})$ gives us $f(a) + f(\frac{1}{a}) \le 2f(1)$, and as $f$ is positive, we obtain that it is bounded. We will show by mathematical induction that for any arbitrary $a \in \mathbb{R}^+$
$$f(a^{2^n}) \le 2^n(f(a) - f(1)) + f(1). \quad (*)$$
We obtain the base case by directly evaluating $P(a^2, 1)$, which yields $f(a^2) \le 2f(a) - f(1) = 2(f(a) - f(1)) + f(1)$. Assume that the statement holds for some $n = k - 1$. From $P(a^{2^k}, 1)$, we obtain $f(a^{2^k}) \le 2f(a^{2^{k-1}}) - f(1)$. From the inductive hypothesis, we have that $f(a^{2^{k-1}}) \le 2^{k-1}(f(a) - f(1)) + f(1)$. By chaining the inequalities we obtain $f(a^{2^k}) \le 2^k(f(a) - f(1)) + f(1)$, which we needed to show. Assume that there exists an $a$ such that $f(a) < f(1)$. As (*) holds true for any arbitrary $a \in \mathbb{R}^+$, we obtain that for a large enough $n$ we will have that $f(a^{2^n}) \le 2^n(f(a) - f(1)) + f(1) < 0$, a contradiction with the fact that our function is positive. Therefore, $f(a) \ge f(1)$ for all $a \in \mathbb{R}^+$. Assume that for some $a$ we have that $f(a) > f(1)$. Since we have that $f(\frac{1}{a}) \ge f(1)$, revisiting $P(a, \frac{1}{a})$ we obtain that $2f(1) < f(a) + f(\frac{1}{a}) \le 2f(1)$, a contradiction. We obtain that the equality $f(a) = f(1)$ must hold true for all $a$, but this contradicts the assumption that $f$ is non-constant.