$f : \mathbb{R}^+ → \mathbb{R}$ with $f(0) = f'(0) = 0$ and $f(x) < x^2$ and $f',f'',f''' > 0$?

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

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.

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 $$f'(x) = \int_0^x f''(t) dt < \int_0^x \frac{\pi }{2} dt = \frac{\pi x}{2}$$ $$f(x) = \int_0^x f'(t) dt < \int_0^x \frac{\pi t}{2} dt = \frac{\pi x^2}{4}$$