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]

Question

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?

Answer 1

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.

You may refer to https://en.wikipedia.org/wiki/Weierstrass_function

Answer 2

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 $x$.