# Quod Erat Demonstrandum

## 2018/03/22

### a question about inequality with derivatives

Filed under: Fun,mathematics,NSS,Pure Mathematics — johnmayhk @ 3:46 下午
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Question

Let $p(x)$ be a polynomial with real coefficients. Prove that if $p(x)-p'(x)-p''(x)+p'''(x)\ge 0$ for any real $x$, then $p(x) \ge 0$ for any real $x$.

Solution (elementary)

$p(x)-p'(x)-p''(x)+p'''(x)\ge 0$

$\Rightarrow (p(x)-p''(x))-(p(x)-p''(x))'\ge 0$

$\Rightarrow q(x)-q'(x)\ge 0$ where $q(x)=p(x)-p''(x)$

$\Rightarrow q(x)\ge q'(x)$ for any real $x$

Claim that $q(x)\ge 0$ for any real $x$.

Proof

Suppose $q(x_0) < 0$, by the condition $q(x)\ge q'(x)$, we have $q'(x_0) < 0$.

That means $q(x)$ is strictly decreasing at $x_0$. And it means that when there is an $x_0$ such that $q(x_0)$ is negative, then the graph of $y=q(x)$ is below the x-axis afterwards ($x > x_0$).

There are 3 cases about the graph of $y=q(x)$.

Case 1: the graph cuts the x-axis

By the above argument, if the graph cuts the x-axis, it cuts once only.

Also, $q(x)$ is a polynomial, we have

$\displaystyle \lim_{x\rightarrow -\infty}q(x)=+\infty$

$\displaystyle \lim_{x\rightarrow +\infty}q(x)=-\infty$

Thus, the degree of $q(x)$, $n$, is odd; and the leading term, $a_nx^n$, is dominant for large $x$; and hence $a_n < 0$.

Now, compare $a_nx^n$ with $na_nx^{n-1}$, the leading term of $q'(x)$.

For $x > n$　（say）

$x^n > nx^{n-1}$

$a_nx^n < na_nx^{n-1}$　（$a_n < 0$

Also, leading terms will be dominant for large $x$, thus, there exist a large enough $x_1$ such that

$q(x_1) < q'(x_1)$

which contradicts that “　$q(x)\ge q'(x)$ for any real $x$　".

Case 2: the graph is below the x-axis (may touch the x-axis at one point)

In this case, the degree of $q(x)$ should be even with leading coefficient $a_n < 0$.

Similar to the argument above, obtaining that

there exist a large enough $x_1$ such that

$q(x_1) < q'(x_1)$

which contradicts that “　$q(x)\ge q'(x)$ for any real $x$　".

Case 3: the graph is above the x-axis (may touch the x-axis at one point)

This is the remaining possible case and it must be true.

That is, $q(x) \ge 0$ for any real $x$, the claims is proved; and the degree of $q(x)$ is even with positive leading coefficient.

Recall that $q(x)=p(x)-p''(x)$; hence $p(x)$ has even degree with positive leading coefficient.

Thus,

$\displaystyle \lim_{x\rightarrow -\infty}p(x)=+\infty$

$\displaystyle \lim_{x\rightarrow +\infty}p(x)=+\infty$

Hence, the polynomial $p(x)$ attains its minimum value at certain value $x=x_2$.

Therefore, $p''(x_2) \ge 0$ ……………….. (*)

But, from above,

$q(x) \ge 0 \Rightarrow p(x) \ge p''(x)$ for any real $x$.

If $p(x_2) < 0$, then $p''(x_2) < 0$ which contradicts to (*).

Therefore $p(x) \ge 0$ for any real $x$.

Q.E.D.