Let, $P(x,y) \Longrightarrow f (x+y)\ge f (x)+yf (f (x))$ We have, $f(x+y) > f(x)$ since $yf (f (x))>0$ So, $f$ is strictly increasing. Again, $P(x,1) \Longrightarrow f(x+1) \ge f(x)+f(f(x)) > f(f(x))$. So, since $f$ is strictly increasing, $x+1 > f(x)$ Now, let, $a=f(f(c))$, $b=f(c)-cf(f(c))$ for some fixed $c$. $P(c,x-c) \Longrightarrow f(x) \ge f(c) + (x-c)f(f(c)) = ax+b$ So, $P(\frac{x}{2}, \frac{x}{2}) \Longrightarrow f(x) > \frac{x}{2}f(f(\frac{x}{2})) \ge \frac{x}{2}(af(\frac{x}{2})+b) \ge \frac{x}{2}(a(a\frac{x}{2}+b)+b) = px^2 + qx + r$, for some reals $p,q,r$ with q positive. Then, $x+1 > f(x) \ge px^2 + qx+r$ Which is definitely not true for sufficiently large $x$. So, NO such function exists!