The function $f(x)$ is strict concave, strict increasing, and $f(0) = 0$. $a, b \in \mathbf R \; and \; a < b$, how can I get that $\frac{a}{b} < \frac{f(a)}{f(b)}$?
Thank you!
Oh sorry I forgot to mention that $f:\mathbf R_{+} \rightarrow \mathbf R_{+}$ is continuous; and $a, b$ are positive...
By definition of concave, assuming $0\lt a\lt b$, $$ f(a)\gt f(0)\frac{b-a}{b}+f(b)\frac ab $$ Divide by $f(b)$ (which is positive by hypothesis): $$ \frac{f(a)}{f(b)}\gt\frac{a}{b} $$
$\frac{f(a)}{f(b)}-\frac{a}{b}=\frac{bf(a)-af(b)}{bf(b)}$. Now, $bf(b)>0\forall b\in \mathbb{R}$ since $f(\cdot)$ is increasing and $f(0)=0$. Now, $$bf(a)-af(b)=(b-a)f(a)-a(f(b)-f(a))$$ For concave functions, if $f$ is differentiable, $\forall x,y\in \mathbb{R}$ $$f(y)\le f(x)+(y-x)f'(x)$$ So, $$f(b)-f(a)<(b-a)f'(a)\\ -f(a)=f(0)-f(a)<-af'(a)$$ Hence, $$bf(a)-af(b)>(b-a)(f(a)-af'(a))>0$$
