# $P(x)+P'''(x)\geq P'(x) + P''(x)$ 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.

• I suppose that by $n$ you mean the degree of the polynomial function. For $n=0$, $P_0(x)=c\in\mathbb{R}$ so if $c\geq 0$ then the conclusion $c>0$ does not necessarily hold. – iolig Oct 20 '16 at 18:01
• Edited it. My bad, $P(x)\geq 0$ – J. Doe Oct 20 '16 at 18:02

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 $P\geq 0$.

• Thank you, very interesting solution. I was looking for a simpler solution, but otherwise a very nice solution. – J. Doe Oct 20 '16 at 19:04
• I think this is exactly the intended solution to this problem. Very nice. – Sungjin Kim Oct 20 '16 at 19:10
• We can have more information such as $P$ has to be of even degree. – Sungjin Kim Oct 20 '16 at 19:11