$f$ is continuous on $[0,\infty)$, differentiable $(0,\infty)$, $f(0)\geq0$ and $f'(x)-f(x)\geq0$. Prove that $f(x)\geq0$ This question was asked in my real analysis quiz and I was unable to solve it. It has 2 parts and unfortunately, I couldn't solve both.

Question:(a) Suppose $f$ is continuous on $[0,\infty) $, differentiable on $(0,\infty)$ and $f(0)\geq0$ . Suppose $f'(x) \geq f(x)$ for all $x \in (0,\infty)$. Show that $f(x)\geq 0$ for all $x \in (0,\infty)$ .

(b) Let $f$ and $g$ are continuous functions on $[0,1]$ satisfying $f(x)\geq g(x)$ for every $0\leq x \leq 1$ , and if $\int_0^{1} f(x) dx =\int_0^{1} g(x) dx $ then show that $f=g$ .
Attempt: In (a) I thought of integrating $f'(x)\geq f(x)$  from $0$ to $x$ with variable t to get $f(x) \geq \int_{0}^{x} f(t) dt  +f(0)$ . But I dont know how to proceed from here.
(b) Continuous functions are integrable so $\int_0^{1} f(t) dt \geq \int_0^{1} g(t) dt  > I$ tried by thinking of teaking sets $A =\{x : f(x)>g(x)\}$ and $B =\{ x: f(x)<g(x) \}$ but could not  proceed .
Kindly Help !!
 A: Consider the function $h:[0,\infty)\rightarrow\mathbb{R}$ defined by $h(x)=e^{-x}f(x)$. Then $h$ is differentiable in $(0,\infty)$. By hypothesis we get $$h'(x)=e^{-x}(f'(x)-f(x))\geq0$$ Then the function $h$ is increasing and hence $$h(x)\geq h(0)=f(0)\geq0\tag{for all $x$}$$$$e^{-x}f(x)\geq0\tag{for all $x$}$$$$f(x)\geq0\tag{since $e^{-x}>0$ for all $x\in[0,\infty)$}$$
Proof of part (b): Use the following lemma to conclude.
Lemma: Let $h:[0,1]\rightarrow\mathbb{R}$ be a non-negative continuous function. Then we have $\int_{0}^{1}h(x)dx=0\iff h\equiv0$.
Proof: It's easy to see that $h\equiv0\implies\int_{0}^{1}h(x)dx=0$. Let, $h\not\equiv0$. Then $\exists$ $t\in[0,1]$ such that $h(t)>0$. Since $h$ is continuous, $\exists$ an $\varepsilon>0$ such that $h(x)>0$ for all $x\in(t-\varepsilon,t+\varepsilon)$. Then we have $$\int_{0}^{1}h(x)dx=\int_{0}^{t-\varepsilon}h(x)dx+\int_{t-\varepsilon}^{t+\varepsilon}h(x)dx+\int_{t+\varepsilon}^{1}h(x)dx\geq\int_{t-\varepsilon}^{t+\varepsilon}h(x)dx>0$$ This completes the proof.
