Could somebody help me to answer the following question?


Let $f:R\_+ \rightarrow R\_+$ 
be a nonindentically zero, nondecreasing, continuous, concave function with $f(0)=0$. Do we have that for any $s,t \in [0,1]$,


$$f(x)f(stx)\leq f(sx)f(tx), \quad \forall x \geq 0.$$


or equivalently, do we have that for any $t\in (0,1)$, $\frac{f(x)}{f(tx)}$ is nonincreasing on $x >0$.


Thanks!



An inequality on concave functions
The answer is no.


Mikael de la Salle has given a counterexample. In a more general setting, the necessary and sufficient condition for the inequality to hold is $g = \log \circ f \circ \exp$ is concave. The inequality can be rewritten into $g(y) + g(a + b + u) \le g(a + y) + g(b + y)$, where $y = \log x$, $a = \log s \le 0$ and $b = \log t \le 0$. This is precisely the necessary and sufficient condition for concavity of $\log \circ f \circ \exp$, which is certainly not the same as concavity of $f$.


