Find $f \in C^1$ when $\forall x\in\mathbb{R}\;h\in\mathbb{R}\;\; f(x) \geq x+1,\: f(x+h) \geq f(x)f(h)$ $$\forall x\in\mathbb{R}\;h\in\mathbb{R}\;\; f(x) \geq x+1,\: f(x+h) \geq f(x)f(h)$$


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*$f(x)$ is defined and differentiable in $\mathbb R$
I assume that $f(x)=e^x$, but I can't prove why...
 A: First note that for each $x \in \mathbb R$ we must have 
$$f(x) \geq f(\frac{x}{2})f(\frac{x}{2}) \geq 0$$
By induction, for all $h >-1$ and $n \in \mathbb N$ you have
$$\frac{f(x+nh)}{f(h)^n} \geq f(x).$$
In particular, for all $a  \in \mathbb R,n> |a|$ and $x$, since $f(x+a)\geq 0$, we have
$$f(x) \leq \frac{f(x+a)}{f(\frac{a}{n})^n} \leq \frac{f(x+a)}{(1+\frac{a}{n})^n}$$
Taking the limit when $n$ foes to infinity, you get that for all $x, a \in \mathbb R$ we have
$$f(x) \leq \frac{f(x+a)}{e^a}$$
This gives
$$f(x) e^a \leq f(x+a)$$
Replacing $x\to x+a, e \to -a$ you get
$$f(x+a) e^{-a} \leq f(x)$$
Therefore, for all $x , a \in \mathbb R$ you get 
$$f(x+a)=e^af(x)$$
Just set $x=0$ to deduce that $f(x)=Ce^x$. Figure out $C$ from $f(x) \geq 1+x$.
Note This solution doesn't uses the differentiability of $f$, which means either I am doing a stupid mistake at some point, or more likely, the solution the book/instructor had in mind is different.
P.S. Added at the beginning why $f \geq 0$, as it was used at some point in thye proof.
