Annoying Polynomial Inequality

Suppose we have a polynomial satisfying $p+p''' \geq p'+p''$ for all $x$. Then prove that $p(x)\geq 0$ for all $x$.

I've been stuck on this problem for weeks. The best I can do is supposing there exists $x$ so that $p(x) < 0$ (for a contradiction), that we can find a point where $p'=0$, $p''>0$, and $p'''>0$, but I can't use this (nor think of anything new) to show it violates the inequality!

Is there a line of thinking I am missing, or am I only missing something little? Is there a better way to go about proving this?

• maybe try rearranging into $(p''' - p'') \geq (p' - p)$ to get some symmetry in the picture? – Scaramouche Jul 31 '13 at 16:43
• Another thing to note is that $q:=p+p'''-p'-p''$ has the same degree and leading coefficient as $p$; this shows you that $p$ must be of even degree and have positive leading coefficient. – Nick Peterson Jul 31 '13 at 17:02
• Why does this show $p$ is of even degree and has positive leading coefficient? – Spine Feast Aug 2 '13 at 21:05

The statement can be slightly stronger:

Claim: If a real polynomial $p$ satisfies that $p+p'''\ge p'+p''$, then either $p>0$ or $p\equiv 0$.

Proof 1: Define $$f(x)=e^x\left(p(x)-2p'(x)+p''(x)\right).$$ Then $\lim\limits_{x\to-\infty}f(x)=0$ and $$f'(x)=e^x\left(p(x)-p'(x)-p''(x)+p'''(x)\right)\ge 0,$$ i.e. $f$ is increasing. It follows that $f\ge 0$, and hence $$p-2p'+p''\ge 0.$$ Define $$g(x)=e^{-x}\left(p(x)-p'(x)\right).$$ Then $\lim\limits_{x\to+\infty}g(x)=0$ and $$g'(x)=-e^{-x}\left(p(x)-2p'(x)+p''(x)\right)\le 0,$$ i.e. $g$ is decreasing. It follows that $g\ge 0$, and hence $$p-p'\ge 0.$$ Define $$h(x)=e^{-x}p(x).$$ Then $\lim\limits_{x\to+\infty}h(x)=0$ and $$h'(x)=-g(x)\le 0,$$ i.e. $h$ is decreasing. Therefore, either $h>0$ or $h\equiv 0$. The conclusion follows.

Proof 2: Denote $q=p-p'$. Then $$p+p'''\ge p'+p''\iff q-q''\ge 0.$$ It implies that $\deg(q-q'')=\deg q=\deg p$ must be even. Therefore, $q$ has a global minimum value point, say $x_q$. Then $$q(x)\ge q(x_q)\ge q''(x_q)\ge 0,$$ i.e. $$p\ge p'.$$ Similarly, $p$ also has a global minimum value point, say $x_p$. Then $$p(x_p)\ge p'(x_p)= 0.$$

If $p(x_p)>0$, we are done. To complete the proof, suppose that $p(x_p)=0$ and let us show that $p\equiv 0$ by reduction to absurdity. If $p(x_p)=0$ and $p$ is not constant, then there exists an integer $k\ge 1$, such that $p(x)=(x-x_p)^k r(x)$ for some polynomial $r$ with $r(x_p)\ne 0$. Since $x_p$ is a minimum point of $p$, $k$ must be even. Then $$p(x)-p'(x)=-kr(x_p)(x-x_p)^{k-1}+O((x-x_p)^k)$$ cannot be always non-negative, a contradiction.

• @GitGud: Since $f$ is increasing and $f(-\infty)=0$, $f\ge 0$. – 23rd Jul 31 '13 at 19:42
• Of course, thanks. – Git Gud Jul 31 '13 at 19:44
• @GitGud: You are welcome. – 23rd Jul 31 '13 at 19:47
• @23rd: From $h$ decreasing and $\lim _{x \to \infty} h(x)=0$ ,how do you conclude either $h \equiv 0$ or $h>0$ ? I seem to only conclude $h \ge 0$ .... – Souvik Dey Jan 3 '15 at 12:41
• @SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial. – 23rd Jan 6 '15 at 16:34