TITLE: Existence of a function
QUESTION [10 upvotes]: I came across this question

Does there exist a function $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $f(x+y)>f(x)(1+yf(x))$ and $x,y\in \mathbb{R}^+$

and I didn't know how to begin on it.

REPLY [4 votes]: Another approach: Rewrite the inequality as $(*): f(x+y)-f(x)>yf(x)^2$. So we see that $f$ is a strictly increasing function. Fix $x$ and let $y\rightarrow+\infty$, we also see that $f$ is unbounded. Hence we can pick a number $a$ such that $f(a)>1$.
Now, for any $x\ge 0$, define $g(x)=f(x+a)-f(a)$. Clearly $g(0)=0$ and $g$ is strictly increasing. Also, it is easy to show that $(**): g(x+y) - g(x)>yg(x)^2$ whenever $x\ge 0$ and $y>0$. So, the inequalities for $f$ and $g$ are similar, except that we allow $x=0$ in the inequality for $g$. However, the function $g$ has a nicer property -- by putting $x=a$ in $(*)$, we have $g(y)>y$ for all $y>0$.
Let $b>0$ and $h=b/n$. By $(**)$, we get $g(ih)-g((i-1)h)>hg((i-1)h)^2$. Summing up for $i=1,2,...,n$ and let $h\rightarrow 0$, we see that $g(b) >\int_0^b g(u)^2 du$. As $g$ is an increasing function, the Riemann integral exists. Now, apply the inequality $g(u)>u$ to the integrand, we get
$$g(b)>\int_0^b g(u)^2 du > \int_0^b u^2 du = \frac{b^3}{3}.$$
Apply this to the integrand, we get
$$g(b)>\int_0^b g(u)^2 du > \int_0^b \left(\frac{u^3}{3}\right)^2 du = \frac{b^7}{3^2 \times 7}.$$
Continue in this manner, we get
$$
g(b) > \frac{b^3}{3},\ \frac{b^7}{3^2 \times 7},\ \frac{b^{15}}{3^4 \times 7^2 \times 15}, ...
$$
The $n$-th term of the above sequence is of order $O(b^{2^{n+1}}/2^{n^3})$. Therefore, when $b>1$, the value of $g(b)$ will eventually blow up, i.e. $f(a+b)$ does not exist.
The final step above suggests that when the domain of $f$ is restricted to a small neighbourhood of zero, $g$ may not blow up [edit: or even the neighbourhood is so small that $a$ and hence $g$ are not defined] and hence a solution for $f$ can exist. For example, if $f$ is only defined on some $(0, b)$, we may take $f(x)=e^{\alpha x}$ as a solution provided that $b$ is small and $\alpha > e^{\alpha b}$.