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Suppose that $f$ be a real valued function that both $f',f''$ exist and satisfy in the following conditions:

  1. $f(0)=0$ and $f'(0)>0$,

  2. For all $x≥0$ , $f''(x) ≥f(x)$,

We want to prove that for all $x>0,~~f(x)>0 $.

Thanks for any hints.

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my last question is different please pay attention to its – Maisam Hedyelloo Jan 15 '13 at 18:31
Is this an old contest problem, or is the contest ongoing? – user53153 Jan 15 '13 at 19:07
this contest problem is for 10 years ago.and this contest math hold every year (i solving these problem for national contest math that will hold this year) – Maisam Hedyelloo Jan 15 '13 at 19:14
The hint given in the answer below is good. To add a little detail, define $b=\inf\{x>0:f(x)=0\}$ and consider the behavior of $f$ on the interval $[0,b]$. – user53153 Jan 15 '13 at 19:17
Are you familiar with the relation between the sign of a derivative and the monotonicity of a function? – 1015 Feb 14 '13 at 14:51
up vote 2 down vote accepted

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.

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My God, that's hard to read. – MyUserIsThis Feb 14 '13 at 14:58
@MyUserIsThis, It's not hard to read, you are not patient to read explained solutions! – AmirHosein SadeghiManesh Feb 14 '13 at 15:02
I am, but some parragraph separation would be appreciated. That's why it's hard, not because it's explaines. Also so many inline equations are not great either. Main equations should be centered in their own line. You won't find math books written like that. – MyUserIsThis Feb 14 '13 at 15:36
@MyUserIsThis, Ok, Perhaps I do it.But I have no enough time to editing. – AmirHosein SadeghiManesh Feb 14 '13 at 18:43
@AmirHosein SadeghiManesh :thanks your approach is nice – Maisam Hedyelloo Feb 14 '13 at 19:27

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?

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