Let $f(x)$ be a polynomial with real coefficients of degree $n$ such that $f(x)$ is non-negative for all real numbers $x$. Define the polynomial $g(x) := f(x) + f'(x) + ... + f^{(n)}(x).$ Show that $g(x)$ is non-negative for all real numbers $x$.
Trivial case $f(x)$ is constant gives $f(x) = g(x)$. If $f(x)$ is not constant, we use that a sum of polynomials of finite degree gives a new polynomial of finite degree, so $g(x)$ is a polynomial of degree $n$ (by construction). For large enough $x$ the highest order polynomial term dominates, since $f(x) \geq 0$ we must have that $g(x) \xrightarrow[]{}\infty, x\xrightarrow[]{}\pm \infty$.
Since $f(x)$ have the same highest order term, otherwise $f(x)$ would be negative for some $x$. This means that there exists an interval $[-R,R]$ such that for all $c > 0$, $g(x) > c$ when $|x| > |R|$. But $g(x)$ is a continuous function and on a bounded set a continuous function attains a minimum (Extreme value theorem).
Assume that $g(x)$ is negative at this minimum, at the minimum we have that $g'(x) = 0$ but
$$ g'(x) = f'(x) + f''(x) + ... + f^{(n)}(x). $$
so
$$ g(x) = f(x) + g'(x) \implies g(x) = f(x) $$
but $f(x)$ was supposed to be positive for all $x$, hence the minimum value of $g(x)$ must be positive.
Thank you to Subhadeep Pal, Bangalore (India), Daniel Holst (Sweden), Jorge Solloa (Mexico City), Anthony (California), B Sashi Kanth (Hyderabad, India), jc (New York), Han Geurdes (Netherlands), and Han Geurdes for submitting solutions to this week’s challenge problem.