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Suppose that $P(x) = a_m x^m + \dots + a_0$ and $Q(x) = b_n x^n + \dots + b_0$ are two polynomials, with $m > n > 1$ and $a_m > b_n > 0$. Suppose that $P$ has $m$ distinct real roots $y_1<\dots<y_m$ and $Q$ has $n$ distinct real roots $z_1<\dots<z_n$.

Is the following claim true: $P(x) - Q(x)$ is strictly increasing for all $x \geq \max \{y_m, z_n\}$.

Note: I have a missing step in a larger proof, which would be fixed by a result like this. In my particular application, I have a bit more specific polynomials and with every numerical example I can come up with the result seems to hold, but I think this result should be more general.

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    $\begingroup$ Try $P(x) = 2 x^3 + 3$, $Q(x) = x^2$. Neither $P$ nor $Q$ has positive roots, but $P-Q$ is decreasing on $(0,1/3)$. $\endgroup$
    – Robert Israel
    Commented Aug 5, 2016 at 19:53
  • $\begingroup$ Thanks for the example. In my application, both polynomials have only real (and in fact $m$ and $n$ distinct) roots. I updated the question. $\endgroup$
    – TomH
    Commented Aug 5, 2016 at 20:31
  • $\begingroup$ OK, try $P(x) = 2 x (x+1)(x+1/3)$ and $Q(x) = x (x+1)$. Note that $P'(0)-Q'(0) = -1/3$. $\endgroup$
    – Robert Israel
    Commented Aug 5, 2016 at 21:09
  • $\begingroup$ Ok, thanks! I think that this proves that my conjecture was wrong. I'll have to find another angle. If you want to, you can state it as an answer, so that I can approve it. $\endgroup$
    – TomH
    Commented Aug 5, 2016 at 21:14

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Try $P(x) = 2 x (x+1)(x+1/3)$ and $Q(x) = x(x+1)$. Note that $P'(0) - Q'(0) = -1/3$.

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