Prove $f(x)\geq e^x$ 
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable function, such that $f^{\ '}\geq f$ and $f(0)=1.$ Prove $f(x)\geq e^x, \forall x\geq0$.

So I started by defining a function $g(x)=f(x)-e^x$, and my intuition tells me that $f^{\ '}\geq f$ means $f$ is increasing, but I'm not sure how to continue.
Any help appreciated.
 A: Consider $g(x)=f(x)e^{-x}$,note that $g$ is increasing and $g(0)=1$ hence $f(x) \geq e^x$ QED
A: I have a semi formal proof to help you along your line of thought.
Let $g(x) = f(x) - e^x$ (as you defined.
Then $g'(x) = f'(x) - e^x$
Clearly $\forall x\ge 0, g'(x) \ge g(x)$ since $f' \ge f$.
Let us look at the zeroes of $g(x)$. Let $a$ be a zero. That is
$g(a) = 0, a\gt 0$ (if there exists no such a, then our proof is already done - justify yourself why)
Then,$g'(a) \ge 0$ but, we know that $g(x)$ is positive in the neighbourhood of x = 0(again justify why)
So, for $g(x)$ to be negative in the neighbourhood of $a$, the derivative must be negative(justify why). But clearly,that is not so. You can also justify from the last inequality that at all further roots, the derivative is only zero at those point(touchig the axis). Hence your proof is done, as $g(x)$ is non negative. Hope this helped.
A: The crux of the proof is the transformation
$$f'(x)\ge f(x)\implies f'(x)e^{-x}-f(x)e^{-x}\ge 0\implies(f(x)e^{-x})'\ge 0.$$
Then $f(x)e^{-x}$ is a non-decreasing function such that $f(0)=1$ and $f(x)e^{-x}\ge1$ for positive $x$.
