Question
Let ![p(x)]( "p(x)")
be a polynomial with real coefficients. Prove that if ![p(x)-p'(x)-p''(x)+p'''(x)\ge 0]( "p(x)-p'(x)-p''(x)+p'''(x)\ge 0")
for any real ![x]( "x")
, then ![p(x) \ge 0]( "p(x) \ge 0")
for any real ![x]( "x")
.
Solution (elementary)
![p(x)-p'(x)-p''(x)+p'''(x)\ge 0]( "p(x)-p'(x)-p''(x)+p'''(x)\ge 0")
![\Rightarrow (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]( "\Rightarrow q(x)-q'(x)\ge 0")
where ![q(x)=p(x)-p''(x)]( "q(x)=p(x)-p''(x)")
![\Rightarrow q(x)\ge q'(x)]( "\Rightarrow q(x)\ge q'(x)")
for any real ![x]( "x")
Claim that ![q(x)\ge 0]( "q(x)\ge 0")
for any real ![x]( "x")
.
Proof
Suppose ![q(x_0) < 0]( "q(x_0) < 0")
, by the condition ![q(x)\ge q'(x)]( "q(x)\ge q'(x)")
, we have ![q'(x_0) < 0]( "q'(x_0) < 0")
.
That means ![q(x)]( "q(x)")
is strictly decreasing at ![x_0]( "x_0")
. And it means that when there is an ![x_0]( "x_0")
such that ![q(x_0)]( "q(x_0)")
is negative, then the graph of ![y=q(x)]( "y=q(x)")
is below the x-axis afterwards (![x > x_0]( "x > x_0")
).
There are 3 cases about the graph of ![y=q(x)]( "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)]( "q(x)")
is a polynomial, we have
![\displaystyle \lim_{x\rightarrow -\infty}q(x)=+\infty]( "\displaystyle \lim_{x\rightarrow -\infty}q(x)=+\infty")
![\displaystyle \lim_{x\rightarrow +\infty}q(x)=-\infty]( "\displaystyle \lim_{x\rightarrow +\infty}q(x)=-\infty")
Thus, the degree of ![q(x)]( "q(x)")
, ![n]( "n")
, is odd; and the leading term, ![a_nx^n]( "a_nx^n")
, is dominant for large ![x]( "x")
; and hence ![a_n < 0]( "a_n < 0")
.
Now, compare ![a_nx^n]( "a_nx^n")
with ![na_nx^{n-1}]( "na_nx^{n-1}")
, the leading term of ![q'(x)]( "q'(x)")
.
For ![x > n]( "x > n")
（say）
![x^n > nx^{n-1}]( "x^n > nx^{n-1}")
![a_nx^n < na_nx^{n-1}]( "a_nx^n < na_nx^{n-1}")
（![a_n < 0]( "a_n < 0")
）
Also, leading terms will be dominant for large ![x]( "x")
, thus, there exist a large enough ![x_1]( "x_1")
such that
![q(x_1) < q'(x_1)]( "q(x_1) < q'(x_1)")
which contradicts that “ ![q(x)\ge q'(x)]( "q(x)\ge q'(x)")
for any real ![x]( "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)]( "q(x)")
should be even with leading coefficient ![a_n < 0]( "a_n < 0")
.
Similar to the argument above, obtaining that
there exist a large enough ![x_1]( "x_1")
such that
![q(x_1) < q'(x_1)]( "q(x_1) < q'(x_1)")
which contradicts that “ ![q(x)\ge q'(x)]( "q(x)\ge q'(x)")
for any real ![x]( "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]( "q(x) \ge 0")
for any real ![x]( "x")
, the claims is proved; and the degree of ![q(x)]( "q(x)")
is even with positive leading coefficient.
Recall that ![q(x)=p(x)-p''(x)]( "q(x)=p(x)-p''(x)")
; hence ![p(x)]( "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")
![\displaystyle \lim_{x\rightarrow +\infty}p(x)=+\infty]( "\displaystyle \lim_{x\rightarrow +\infty}p(x)=+\infty")
Hence, the polynomial ![p(x)]( "p(x)")
attains its minimum value at certain value ![x=x_2]( "x=x_2")
.
Therefore, ![p''(x_2) \ge 0]( "p''(x_2) \ge 0")
……………….. (*)
But, from above,
![q(x) \ge 0 \Rightarrow p(x) \ge p''(x)]( "q(x) \ge 0 \Rightarrow p(x) \ge p''(x)")
for any real ![x]( "x")
.
If ![p(x_2) < 0]( "p(x_2) < 0")
, then ![p''(x_2) < 0]( "p''(x_2) < 0")
which contradicts to (*).
Therefore ![p(x) \ge 0]( "p(x) \ge 0")
for any real ![x]( "x")
.
Q.E.D.
