# Let $f:\mathbb R^+\to\mathbb R$ be a continuous function satisfied $f(a)+f(b)\ge f(2\sqrt{ab})$ for all $a,b>0$ , is $f$ differentiable? [closed]

Let $$f:\mathbb R^+\to\mathbb R$$ be a continuous function satisfied $$f(a)+f(b)\ge f(2\sqrt{ab})$$ for all $$a,b>0$$ , is $$f$$ differentiable?

Morever, if for all $$a_1,a_2,\cdots,a_n>0$$ there holds $$\sum_if(a_i)\ge f\left(n\sqrt[n]{\prod_ia_i}\right)\\$$ is $$f$$ differentiable?

• What have you tried?
– Joe
Jun 16, 2022 at 4:07
• I think it's not true but I can't find an example. Jun 16, 2022 at 4:10
• Hint: Let $a = e^{x / 2}$ and $b = e^{y / 2}$. Jun 16, 2022 at 4:16
• Letting $b=4a$ tells you that $f(a) \ge 0$. Jun 16, 2022 at 4:47
• Please put the main question in the body of the post, not just the title. Jun 16, 2022 at 5:29

Easy example: find $$f$$ such that $$2\leq f(x)\leq 3$$ for all $$x\in\mathbb{R}$$ and $$f$$ is everywhere continuous but nowhere differentiable.
Explicit demonstration: Let $$x\gt0$$ be such that $$f(2x)\neq0$$ and $$f(x)\neq0$$. Define $$y$$ such that $$2x=e^y$$, which is possible by surjectivity. $$f$$ cannot be differentiable at the point $$x$$, since: \begin{align}\liminf_{h\to0}\frac{f(e^{2h}e^y/2)-f(e^y/2)}{(e^{2h}-1)e^{y/2}}&\ge\limsup_{h\to0}\frac{f(e^{y+h})}{(e^{2h}-1)e^y/2}\\&=\infty\end{align}Since by continuity $$f(e^{y+h})\to f(e^y)=f(2x)\neq0$$ but the denominator tends to $$0$$. A similar argument is applicable for negative $$x$$.