Posted before (in several versions) - do a search.

WakeUp wrote: Prove that there is no function $f:(0,\infty )\rightarrow (0,\infty)$ such that \[f(x+y)\ge f(x)+yf(f(x)) \] for every $x,y\in (0,\infty )$. Adaptation of one of the multiple previously posted solutions : Let $P(x,y)$ be the assertion $f(x+y)\ge f(x)+yf(f(x))$ 1) $f(x)< x+1$ $\forall x>0$ =================== If $f(x)\le x$, we get obviously $f(x)< x+1$ If $f(x)>x$ for some $x$, then $P(x,f(x)-x)$ $\implies$ $f(f(x))\ge f(x)+(f(x)-x)f(f(x))>(f(x)-x)f(f(x))$ $\implies$ $f(x)< x+1$ Q.E.D. 2) $f(f(x))\le 1$ $\forall x>0$ ==================== If $f(f(x))>1$ for some $x$, then Let $y>\frac{x+1}{f(f(x))-1}$ so that $yf(f(x))>x+y+1$ : $P(x,y)$ $\implies$ $f(x+y)\ge f(x)+yf(f(x))>x+y+1$ and so contradiction with 1) above. Q.E.D. 3) $\exists M>0$ such that $f(x)<M$ $\forall x>0$ ================================== Suppose that $\forall a>0,\exists u_a>0$ such that $f(u_a)>a$ Let $x>0$ and $a>x$ and $u_a>0$ such that $f(u_a)>a>x$ $P(x,f(u_a)-x)$ $\implies$ $f(f(u_a))\ge f(x)+(f(u_a)-x)f(f(x))>(a-x)f(f(x))$ $\implies$ (using 2) above) $1>(a-x)f(f(x))$ So $0<f(f(x))<\frac 1{a-x}$ $\forall a>x$ which is impossible Q.E.D. 4) No such function exists ================== $P(x,y)$ $\implies$ $M>yf(f(x))$ and so $0<f(f(x))<\frac My$ $\forall y$ which is impossible Q.E.D.

mavropnevma wrote: Posted before (in several versions) - do a search. Apologies for this, I searched but out of the matches that came up, they were either unsolved or slightly different (such as http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=307863).

WakeUp wrote: Prove that there is no function $f:(0,\infty )\rightarrow (0,\infty)$ such that \[f(x+y)\ge f(x)+yf(f(x)) \] for every $x,y\in (0,\infty )$. The problem, as such, was used in some Italian math competition. A strengthened version, for the relation $f(x+y)\ge yf(f(x))$, was asked at the 2009 IMAR (Romanian Institute for Mathematics) competition. Detailed proof(s), generalization, and discussions to be found in the RMC 2009 brochure.