# Suppose we have a polynomial satisfying p+p′′′>=p′+p′′ for all x. Then prove that p(x)>=0 for all x.

Suppose we have a polynomial satisfying $p+{p}^{‴}\ge {p}^{\prime }+{p}^{″}$ for all x. Then prove that $p\left(x\right)\ge 0$ for all x.
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Define
$f\left(x\right)={e}^{x}\left(p\left(x\right)-2{p}^{\prime }\left(x\right)+{p}^{″}\left(x\right)\right).$
Then $\underset{x\to -\mathrm{\infty }}{lim}f\left(x\right)=0$ and
${f}^{\prime }\left(x\right)={e}^{x}\left(p\left(x\right)-{p}^{\prime }\left(x\right)-{p}^{″}\left(x\right)+{p}^{‴}\left(x\right)\right)\ge 0,$
i.e. f is increasing. It follows that $f\ge 0$, and hence
$p-2{p}^{\prime }+{p}^{″}\ge 0.$
Define
$g\left(x\right)={e}^{-x}\left(p\left(x\right)-{p}^{\prime }\left(x\right)\right).$
Then $\underset{x\to +\mathrm{\infty }}{lim}g\left(x\right)=0$ and
${g}^{\prime }\left(x\right)=-{e}^{-x}\left(p\left(x\right)-2{p}^{\prime }\left(x\right)+{p}^{″}\left(x\right)\right)\le 0,$
i.e. g is decreasing. It follows that $g\ge 0$, and hence
$p-{p}^{\prime }\ge 0.$
Define
$h\left(x\right)={e}^{-x}p\left(x\right).$
Then $\underset{x\to +\mathrm{\infty }}{lim}h\left(x\right)=0$ and
${h}^{\prime }\left(x\right)=-g\left(x\right)\le 0,$
i.e. h is decreasing. Therefore, either $h>0$ or $h\equiv 0$. The conclusion follows.