A question around concept of derivative Suppose that $f$ be a real valued function that both $f',f''$ exist and  satisfy in the following conditions:


*

*$f(0)=0$ and  $f'(0)>0$,


*For all $x≥0$ , $f''(x) ≥f(x)$,

We want to prove that  for all $x>0,~~f(x)>0 $.
Thanks for any hints.
 A: I'll tell you idea of proof, you can try to write a rigorous one.
From 1. we can see that $f(x)$ initially goes positive. For $f(x)$ to reach $0$ again, it needs $f'(x)<0$ somewhere. But how can this happen if $f(x)$ is still $\ge 0$ and $2$. is given?
A: Because $f''$ exists in all $\mathbb{R}^{\geq0}$ so $f'$ is continuous in this interval. because $f'$ is continuous and $f'(0)>0$ so there exists $\varepsilon>0$ such that $f'$ is strictly positive in $[0,\varepsilon)$. So we know $\emptyset\neq A:=\{r\in\mathbb{R}^{\geq0}|\;f'\text{is strictly positive in} [0,r)\}\subset\mathbb{R}$. If there exists a positive real number like $a$ out of $A$ then note that for every greater number like $b>a$ if $f'$ be strictly positive on $[0,b)$ then it is same on $[0,a)$ and it is contradiction with $a$ is not in $A$, so $a$ should be an upper bound for $A$. So by assuming $A\neq\mathbb{R}^{\geq0}$ ,$A$ will be nonempty subset of $\mathbb{R}$ which is bounded from above and by Consummate principle of real numbers, $A$ takes its suprimum, say $a$ . Therefore $f'$ is strictly positive on $[0,a)$. If $f'(a)<0$ then by the intermediate theorem for continuous functions and as $f'(\frac{a}{2})>0$, $f'(a)<0$ there should be $c\in[\frac{a}{2},a)\subset[0,a)$ such that $f'(c)=0$ that is contradiction, again if $f'(a)=0$ then we pay attention that because $f''$ exists in whole $\mathbb{R}^{\geq0}$ so $f''(a)=f''(a^{-})=\lim_{x\rightarrow a,\;x<a}\frac{f'(x)-f'(a)}{x-a}\lim_{x\rightarrow a,\;x<a}\frac{f'(x)}{x-a}$, But pay attention that for every $x<a$ we have $x-a<0$ and $f'(x)>0$ which implies $f''(a)<0=f'(a)$ that is contradiction with the question's assumption. Therefore $f'(a)$ should be strictly positive and by continuity of $f'$ there is $\varepsilon>0$ such that $f'$ is strictly positive on $(a-\varepsilon,a+\varepsilon)$ it get us $f'$ is strictly positive on $[0,a+\varepsilon)$ and then $a+\varepsilon\in A$ while $a+\varepsilon>a=sup A$ that is contradiction. So the assumption "there exists a positive real number like $a$ out of $A$ " is invalid. So we showed $f'$ is strictly positive in whole $\mathbb{R}^{\geq0}$. This means $f$ is strictly increasing in this interval and because $f'$ exists in this interval so $f$ is continuous and because $f(0)=0$ and $f$ is strictly increasing we have $\forall x>0\; :\; f(x)>f(0)=0$. And this is what you want.
