P(x)+P‴(x)≥P′(x)+P″(x)P(x)+P”'(x)\geq P'(x) + P”(x) then P(x)>0 \, \forall x\in \mathbb R P(x)>0 \, \forall x\in \mathbb R

Let P(x)P(x) be a polynomial function of real coefficients with the following property:
P(x)+P”'(x)\geq P'(x) + P”(x) P(x)+P”'(x)\geq P'(x) + P”(x)
then, P(x)\geq0 \quad \forall x\in \mathbb R 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 nn you mean the degree of the polynomial function. For n=0n=0, P_0(x)=c\in\mathbb{R}P_0(x)=c\in\mathbb{R} so if c\geq 0c\geq 0 then the conclusion c>0c>0 does not necessarily hold.
– ioanna202
Oct 20 at 18:01



Edited it. My bad, P(x)\geq 0P(x)\geq 0
– J. Doe
Oct 20 at 18:02


1 Answer


Hmm, a bit far-fatched answer:

The initial inequality rewrites as P”’-P”\geq P’-PP”’-P”\geq P’-P. Letting Q=P’-PQ=P’-P, this yields Q”\geq QQ”\geq Q.

Adding Q’Q’ on both sides, Q”+Q’ \geq Q’+QQ”+Q’ \geq Q’+Q.

Letting R=Q’+QR=Q’+Q we have R’\geq RR’\geq R. Multiplying by the positive quantity e^{-x}e^{-x} yields R’e^{-x}-Re^{-x}\geq 0R’e^{-x}-Re^{-x}\geq 0

which rewrites as \frac{d}{dx}\left(R e^{-x}\right)\geq 0\frac{d}{dx}\left(R e^{-x}\right)\geq 0

The function x\mapsto Re^{-x}x\mapsto Re^{-x} is therefore increasing. Furthermore, since RR is polynomial, \lim_{\infty }Re^{-x} = 0\lim_{\infty }Re^{-x} = 0.

A continuous, increasing function which goes to 00 at \infty\infty must be \leq 0\leq 0.

Hence Re^{-x}\leq 0Re^{-x}\leq 0 and R\leq 0R\leq 0.

But R = Q’+Q = P”-P’+P’-P = P”-PR = Q’+Q = P”-P’+P’-P = P”-P.

Hence P”\leq PP”\leq P.

Let’s perform the same trick one more time. Letting S= P’+PS= P’+P, the last inequality yields S’\leq SS’\leq S, hence \frac{d}{dx}\left(S e^{-x}\right)\leq 0\frac{d}{dx}\left(S e^{-x}\right)\leq 0.

The functions x\mapsto S e^{-x}x\mapsto S e^{-x} is therefore decreasing, continuous, and goes to 00 at \infty\infty. Hence S e^{-x}\geq 0S e^{-x}\geq 0 and S\geq 0S\geq 0.

This means P’+P\geq 0P’+P\geq 0, hence \frac{d}{dx}\left(P e^{x}\right)\geq 0\frac{d}{dx}\left(P e^{x}\right)\geq 0.

The functions x\mapsto P e^{x}x\mapsto P e^{x} is increasing and goes to 00 at -\infty-\infty.

Hence P e^{x} \geq 0P e^{x} \geq 0 and P\geq 0P\geq 0.



Thank you, very interesting solution. I was looking for a simpler solution, but otherwise a very nice solution.
– J. Doe
Oct 20 at 19:04



I think this is exactly the intended solution to this problem. Very nice.
– i707107
Oct 20 at 19:10



We can have more information such as PP has to be of even degree.
– i707107
Oct 20 at 19:11