Question: > 
> If $p(x)$ is a polynomial with a positive leading coefficient and even degree such that $p(x)\ge p''(x)$ for all real x, then $p(x)\ge 0$ for all real x.
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I tried considering the case where $p(x) < 0$ for some x and limiting behaviour. $p'(x)$ is eventually negative for small enough x and eventually positive for large enough x. Also, p is eventually positive for all x with large enough absolute value. I was thinking of applying the Intermediate value theorem somehow, but I can't seem to get a contradiction.

Answer: 
A real polynomial with even degree and positive leading coefficients attains its *minimum* at some point $x\_0$, and at a minimum is $p''(x\_0) \ge 0$. It follows that
$$
 p(x) \ge p(x\_0) \ge p''(x\_0) \ge 0
$$
for all $x \in \Bbb R$.
Since the polynomial has even degree and the leading coefficient is positive, $p(x)\to +\infty$ as $x\to \pm\infty$, so it will have an absolute minimum at some point $x\_0\in\mathbb{R}$.

Since $p$ is everywhere infinitely differentiable, in particular, $p'(x\_0)=0$ and $p''(x\_0)$ cannot be negative since it would mean that $p$ would have a maximum at $x\_0$. Hence, $p''(x\_0)\geq 0$.

Finally, using the problem hypothesis, $p(x\_0)\geq p''(x\_0)\geq 0$. Since $p(x\_0)$ was the minimum value attained by $p$, $p(x)\geq p(x\_0)\geq 0$ for all $x\in\mathbb{R}$.