Suppose $f$ is a polynomial function with degree $n$ and $f(x)\geqslant0$ (so $n$ must be even). Prove that $f + f' + f'' +\ldots+f^{(n)}\geqslant0$.
Put $\;g(x)=f + f' + f'' +\ldots+f^{(n)}\;$ and $h(x)=f' + f'' +\ldots+f^{(n)}$.
Suppose $g(x) < 0$ for all $x$, then $f < -h(x)$. Since $h(x)$ is odd degree then $h(x)$ contain both negative and positive value implies that $f(x)<0$ for some $x$. Contradict the fact $f \geqslant0$.
Suppose $g(x) < 0$ for some $x$. Which mean $g(x) < 0$ for some $x$ and $g(x) > 0$ for others, then $g(x)=0$ for some $x$. Contradict assumption $g(x)<0$. Hence $g(x)\geqslant0$.
Is that a true proof? Is there any proof which doesn't use contradiction?
Your proof does not work. It seems that you want to give a proof of $g(x) \ge 0$ by contradiction. That is, you want to show that the assumption that $g(x) < 0$ for some $x$ leads to a contradiction.
Case 1. $g(x) < 0$ for all $x$.
This part is correct; you get a contradiction which shows that $g(x) \ge 0$ for some $x$.
Case 2. $g(x_0) < 0$ for some $x_0$, but $g(x_1) > 0$ for some other $x_1$.
I do not see why you consider this case; all you do with it is to deduce the existence of some $x_2$ with $g(x_2) = 0$. What you should do is to investigate the situation that $g(x_0) < 0$ for some $x_0$ and $g(x_1) \ge 0$ for some $x_1$. This also shows that $g(x_2) = 0$ for some $x_2$, but I do not know why you need this fact. Anyway, there is no contradiction here unless you can show that $x_0 = x_1$ or $x_0 = x_2$.
You made a logical mistake when you wrote
The two statements
do not contradict each other. It seems that you think that taking both together is equivalent to
which is not true.