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Now suppose that $f(x) - f'(x) \geqslant0$ for every real $x$. Then $f(x) - f'(x)\to +\infty$ as $x\to\infty$ and therefore $f(x)\to +\infty$ as $x\to\infty$. By the same argument, $f(x)\to +\infty$ as $x\to-\infty$. It follows that $f(x)$ has a finite minimum value, which it attains at some point $x_0$. Then $f'(x_0) = 0$. But $f(x_0) - f'(x_0) \geqslant0$. So $f(x_0)\geqslant0$. But if the minimum value of $f(x)$ is nonnegative then the function must be nonnegative everywhere. That proves

It follows that

To prove Fact 2, let $g(x) = f(-x)$. Then $g'(x) = -f'(-x)$ and so $g(x) - g'(x) = f(-x) + f'(-x) \geqslant0$ for every real $x$. It follows from Fact 1 that $g(x)\geqslant0$ for every real $x$. Therefore $f(x)\geqslant0$ for every real $x$.

Write $D= \frac d{dx}$ for the operator of differentiation, and $I$ for the identity operator. Then those two Facts can be written as

Now let $p(x)$ be a polynomial with real coefficients such that $p(x) - p'(x) - p''(x) + p'''(x) \geqslant0$ for every real $x$. That condition says that $(I-D - D^2 + D^3)p(x)\geqslant0$ for every real $x$. But $I-D-D^2+D^3 = (I-D)(I-D^2)$. Therefore $(I-D - D^2 + D^3)p(x) = (I-D)\bigl((I-D^2)p(x)\bigr) \geqslant0$ for every real $x$. It follows from Fact 1 that $(I-D^2)p(x) \geqslant0$ for every real $x$. But $I-D^2 = (I-D)(I+D)$. Therefore $(I-D)\bigl((I+D)p(x)\bigr) \geqslant0$ for every real $x$, and from Fact 1 again it follows that $(I+D)p(x) \geqslant0$ for every real $x$. Finally, by Fact 2 it then follows that $p(x) \geqslant0$ for every real $x$.