$f : \mathbb{R}^+ → \mathbb{R}$ with $f(0) = f'(0) = 0$ and $f(x) < x^2$ and $f',f'',f''' > 0$? I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function. 
Here is the statement:

There is no function $f(x)$ on $x \ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.

I appreciate any help!
 A: As in the first answer and comment by Daniel, we can choose any $f'''$ that is positive as long as its definite integral out to infinity is finite and smaller than 2. then we back up, each function is the definite integral from $0.$
$$  f'''(x) = \frac{1}{1+x^2} $$
$$  f''(x) = \arctan x $$
$$  f'(x) = x \arctan x - \frac{1}{2} \log \left(1+x^2 \right) $$
$$ f(x) =  \left( \frac{x^2 -1}{2} \right) \arctan x - \frac{x}{2} \log \left(1+x^2 \right) +  \frac{x}{2}$$
As $f'' < \frac{\pi}{2},$ we get $$f'(x) = \int_0^x f''(t) dt <  \int_0^x \frac{\pi }{2} dt =  \frac{\pi x}{2}$$
$$f(x) = \int_0^x f'(t) dt <  \int_0^x \frac{\pi t}{2} dt =  \frac{\pi x^2}{4}$$
A: I do not think it is true.  Take for example 


*

*$f(x) = 1 - x + \frac12x^2 - e^{-x}$

*$f'(x) = - 1 + x + e^{-x}$

*$f''(x) = 1 - e^{-x}$

*$f'''(x) = e^{-x}$
Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases   

A: Your intuition is correct for a slightly different statement:

There is no function $f(x)$ on $x \ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the third derivative of $f(x)$ is bounded below by some $\epsilon > 0$ on $x>0$.

The argument is as follows:  Suppose that such a function exists.  Then we have $f'''(x) \geq \epsilon$, which means that (integrating both sides)
$$
f''(x) - f''(0) \geq \epsilon x.
$$
Denote $f''(0) = a$, so that $f''(x) \geq \epsilon x + a$.  Integrating two more times, we have
$$
f'(x) \geq \frac{1}{2} \epsilon x^2 + a x + f'(0) = \frac{1}{2} \epsilon x^2 + a x
$$ $$
f(x) \geq \frac{1}{6} \epsilon x^3 + a x^2 + f(0) = \frac{1}{6} \epsilon x^3 + \frac{1}{2} a x^2
$$
But if $f(x) < x^2$, we have
$$
x^2 > f(x) \geq \frac{1}{6} \epsilon x^3 + \frac{1}{2} a x^2
$$
for all $x > 0$, which reduces to
$$
\frac{6(1 - a/2)}{\epsilon} > x.
$$
for all $x > 0$.  For any value of $\epsilon > 0$ and $a \in \mathbb{R}$, this will be violated by some $x > 0$, and we have a contradiction.
Intuitionally:  if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $\frac{1}{6} x^3$ times that value.  But any function that grows as fast as $x^3$ (times any positive number) must have a greater value than $x^2$ at some point.
A: Just a comment: R+ is the set of STRICTLY POSITIVE reals, so I think R+ in opening statement needs to be replaced by the interval [0, infinity).
