Prove that there is no function $f:\Bbb{R}\to\Bbb{R}$ with $f(0)>0$ such that $\forall x,y\in\Bbb{R}, f(x+y)\geq f(x)+y f(f(x))$ I got this problem:
Prove that there is no function $f:\Bbb{R}\to\Bbb{R}$ with $f(0)>0$ such that $\forall x,y\in\Bbb{R}, f(x+y)\geq f(x)+y f(f(x))$. 
(Hint: the solution involves limits at infinity)
I tried to prove it (by contradiction) but I failed.
Thanks on any help.
 A: DISCLAIMER: This answer is incorrect as it stands, namely in the $k>0$ section is claims that:
$$
f(0)=f(x-x)\geq f(x)-x f(f(x))
$$
implies $f(x)<f(0)$, which is false.
This disclaimer will stay here until the problem is fixed.

From the condition $f(0)>0$ it follows that with $x=0$ we have
$$
f(0+y)-y f(f(0))\geq f(0)>0
$$
so it follows that $f(y)>ky$ where $k=f(f(0))$ is some non-zero constant. It has to be non-zero since otherwise $0=k=f(f(x))>k f(x)=0$. Now we have to divide into cases.
If $k>0$
If $k>0$ it follows that $x f(f(x))>kx f(x)>k^2 x^2\geq 0$. So in that case we have
$$
\underbrace{f(x-x)}_{f(0)}\geq f(x)-\underbrace{x f(f(x))}_{\text{strictly greater than }0}\\
\iff\\
f(x)<f(0)
$$
But the above statement is absurd by all means. For one thing $f(0)<f(0)$ and moreover $f(0)>f(x)>kx\rightarrow\infty$ which is also absurd.
If $k<0$
If we apply the beginning of what Denis showed, namely that $f(x)\rightarrow\infty$ for $x\rightarrow\infty$, then we get in addition to that from my analysis that for $k<0$ we have $f(-x)\geq -kx\rightarrow\infty$ for $x\rightarrow\infty$. So we see that
$$
f(0)=f(-x+x)\geq f(-x)+x f(f(-x))\rightarrow\infty
$$
for $x\rightarrow\infty$ too. A contradiction since we have just shown that the constant $f(0)$ is greater than or equal to an expression that tends to infinity. So $k<0$ is impossible too.
A: Let $a=f(0)>0$, $b=f(a)$ and $c=f(b)$.
We know that $f(0+b)\geq a+b^2$, so we get $c>0$.
We also have for all $x$, $f(x+a)\geq f(a)+cx$, so in particular $\lim_{x\to\infty} f(x)=\infty$ and more precisely $f(x)=\Omega(x)$ when $x\to\infty$.
This means that for $x$ big enough, $f(x)>0$, and therefore, $f(x+1)\geq f(x)+f(f(x))>f(f(x))$. This gets us $f(f(x-1))<f(x)$.
Moreover, for $x$ big enough, we have $f(x-1)>x$ (since $f(x+1)=\Omega(x^2)$).
Let us choose $x_0$ big enough, and define $x_{n+1}=f(x_n-1)$.
The sequence $(x_n)$ is strictly increasing, and for all $n$, we have $f(x_{n+1})<f(x_n)$. We reached a a contradiction with $\lim_{x\to\infty} f(x)=\infty$.
A: By what Denis said we get that $lim_{x\to\infty}f(x)=\infty$  And so $lim_{x\to\infty}f(f(x))=\infty$.
Now we will reach a contradiction by using the definition of limits at infinity:
Because $lim_{x\to\infty}f(x)=\infty$ and $lim_{x\to\infty}f(f(x))=\infty$. We get that: 
$\forall 0<M_1,\exists 0<N_1$ such that $\forall N_1<x, M_1<f(x)$ and that 
$\forall 0<M_2,\exists 0<N_2$ such that $\forall N_2<x, M_2<f(f(x))$
Now take $M_1=M_2=1$ and we will get that $\exists 0<N_1$ such that $\forall N_1<x, 1<f(x)$ and that $\exists 0<N_2$ such that $\forall N_2<x, 1<f(f(x))$
Take $x=max (N_1+1,N_2+1)$, then $N_1,N _2<x$ and so we get that $1<f(x)$ and that $1<f(f(x))$
Now take $y=max(\frac{x+1}{f( f( x))-1},N_2-x)$ which implies that $N_2< x+y+1\leq y f(f(x ))$ and so $1<f(f(x+y+1))$, Now $f (x+y)\geq f(x )+yf(f(x))\geq x+y+1$ And hence
$f(f(x+y))\geq f(x+y+1)+(f(x+y)-(x+y+1))f(f(x+y+1)) \geq f(x +y+1)\geq f(x+y)+f(f(x+y))\geq f(x)+yf(f(x))+f(f(x+y))>f(f(x+y))$ 
And so we get that $0>0$ which is a contradiction.
