Calculus Challenge III: Prove $f(x)>0$ for All Real $x$

[anemone]

Let $f(x)$ be a polynomial with real coefficients, satisfying $f(x)-f'(x)-f''(x)+f'''(x)>0$ for all real $x$.

Prove that $f(x)>0$ for all real $x$.

 

[anemone]

Hint:

Spoiler

Let $g(x)=f(x)-f''(x)$.

 

[anemone]

Solution of other:

Spoiler

If we put $g(x)=f(x)-f''(x)$, then the given condition can be written $g(x)-g'(x)>0$.

We show first that this implies $g(x)>0$ for all $x$, and then it is easy to show that $f(x)>0$ for all $x$.

That $g(x)-g'(x)>0$ implies that $g(x)-g'(x)$, and therefore $g(x)$ itself, has even degree and positive leading coefficient. Thus $g(x)$ has an absolute minimum at some point, say at $a$. Since $g'(a)=0$, we have $g(x)>g(a)>g'(a)>0$ for all $x$, as claimed.

This implies that $f(x)-f''(x)=g(x)$ is a polynomial of even degree with leading coefficient positive. Then $f(x)$ has an absolute minimum value, say at $x=b$. At a minimum point, the second derivative satisfies $f''(b)>0$. Then for all $x$ we have $f(x)>f(b)>f''(b)>0$.