If $p(x)$ has even degree and a positive leading coefficient and $p(x)\ge p''(x)$, then $p(x)\ge0$

Question

This is my solution, though not too sure of its rigor:

If $p(x)$ has $n$ degree,then we differentiate $n+1$ times, and and this polynomial will equal zero. We then integrate this $n+1$ times using $p(x)\ge p''(x)$, meaning that the integral of the integral is greater or equal to zero, and carry on integrating till we reach $p(x)$, which then must be greater or equal to zero.

AKA $p(x)>p''(x)\cdots>0$

Answer 1

Hints in highlights:

First, be sure you understand why $\;\lim\limits_{x\to\pm\infty}f(x)=+\infty\;$ .

Next: suppose there's $\;a\in\Bbb R\;$ s.t. $\;f(a)<0\;$ . By the above + continuity, differentiability of $\;f(x)\;$ it follows $\;f(x)\;$ has a local minimum at some point where the value of the function is negative.

Now use the given data and get a contradiction.