$P(x)+P'''(x)\geq P'(x) + P''(x)$ then $P(x)>0 \, \forall x\in \mathbb R $ Let $P(x)$ be a polynomial function of real coefficients with the following property:
$$P(x)+P'''(x)\geq P'(x) + P''(x) $$
then, $$P(x)\geq0 \quad \forall x\in \mathbb R $$
I've tried writing the polynomial in its expanded form and cancelling terms, without any real success. Thanks for your help.
 A: Hmm, a bit far-fatched answer:
The initial inequality rewrites as $P'''-P''\geq P'-P$. Letting $Q=P'-P$, this yields $Q''\geq Q$.
Adding $Q'$ on both sides,  $Q''+Q' \geq Q'+Q$.
Letting $R=Q'+Q$ we have $R'\geq R$. Multiplying by the positive quantity $e^{-x}$ yields $$R'e^{-x}-Re^{-x}\geq 0$$
which rewrites as $$\frac{d}{dx}\left(R e^{-x}\right)\geq 0$$
The function $x\mapsto Re^{-x}$ is therefore increasing. Furthermore, since $R$ is polynomial, $\lim_{\infty }Re^{-x} = 0$.
A continuous, increasing function which goes to $0$ at $\infty$ must be $\leq 0$.
Hence $Re^{-x}\leq 0$ and $R\leq 0$.
But $R = Q'+Q = P''-P'+P'-P = P''-P$.
Hence $P''\leq P$.
Let's perform the same trick one more time. Letting $S= P'+P$, the last inequality yields $S'\leq S$, hence $\frac{d}{dx}\left(S e^{-x}\right)\leq 0$.
The functions $x\mapsto S e^{-x}$ is therefore decreasing, continuous, and goes to $0$ at $\infty$. Hence $S e^{-x}\geq 0$ and $S\geq 0$.
This means $P'+P\geq 0$, hence $\frac{d}{dx}\left(P e^{x}\right)\geq 0$.
The functions $x\mapsto P e^{x}$ is increasing and goes to $0$ at $-\infty$.
Hence $P e^{x} \geq 0$ and $P\geq 0$.
