Balkan Mathematical Olympiad Shortlisted Problems

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 $a\in {\mathbb{R}}^{+}$ $$f(a^{2^n})\le 2^n(f(a)-f(1))+f(1).(\ast )$$ 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.