Polynomial inequality: show $p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n\ge 0$

Solution 1:

You can do this by showing that the numerator of

$$(2k+1)p(x)={2kx^{2k+2}+(2k+2)x^{2k+1}-(2k+2)x-2k\over x^2-1}$$

has just two zeroes, at $x=\pm1$ (so that elsewhere it has the same sign as the denominator, since it clearly tends to $+\infty$ as $x\to\pm\infty$ and is negative at $x=0$). This can be done by looking closely at the numerator's first and second derivatives,

$$(2k+2)(2k)x^{2k+1}+(2k+2)(2k+1)x^{2k}-(2k+2)$$

and

$$(2k+2)(2k+1)(2k)x^{2k}+(2k+2)(2k+1)(2k)x^{2k-1}=(2k+2)(2k+1)(2k)x^{2k-1}(x+1)$$

We see that the second derivative is positive to the left of $-1$ and to the right of $0$, and negative in between. The implies the first derivative is increasing on $(-\infty,-1)$, decreasing on $(-1,0)$, and increasing on $(0,\infty)$. Since the value of the first derivative at $x=-1$ is $(2k+2)(-2k+(2k+1)-1)=0$, the first derivative is non-positive on $(-\infty,0)$, so the numerator is strictly decreasing on that interval, passing through $0$ at $x=-1$. On $(0,1)$, the first derivative increases from a negative value at $x=0$ to a positive value at $x=1$ so the numerator decreases to a minimum (negative) value and then increases to pass through $0$ at $x=1$. It continues increasing on $(1,\infty)$, and thus it has just the two zeroes, at $x=\pm1$.

Solution 2:

$$p(x) = \frac{2}{ 2k+ 1} ( k + (k+1) x + k x^2 + (k+1)x^3 + \ldots + (k+1) x^{2k-1} + k x^{2k} ) $$

By considering signs (IE Replace $x$ with $-x$), we WTS that for $ x \geq 0$,

$$ k( 1 + x^2 + \ldots + x^{2k}) \geq (k+1)( x + x^3 + \ldots + x^{2k-1}) .$$

Since the number of terms on each side are equal to $k(k+1)$, and the sum of the degrees are the same, it suggests that we can proceed via smoothing out these terms.

This can by done, say by summing up the following inequalities (overall relevant values of $i$):

• $ \frac{k}{2} x^{2i} + \frac{k}{2} x^{2i+2} \geq k x^{2i+1} $
  • This gives us $ \frac{k}{2}1 + kx^2 + \ldots + kx^{2k-2} + \frac{k}{2} x^{2k} \geq k( x + x^3 + \ldots + x^{2k-1}) .$
• $ \frac{1}{2} x^0 + \frac{1}{2} x^{2k} \geq \frac{1}{2}x^{2i-1 } + \frac{1}{2}x^ { 2k-2i+1} $
  • This gives us $ \frac{k}{2} 1 + \frac{k}{2} x^{2k} \geq ( x + x^3 + \ldots + x^{2k-1})$.

Notes:

• This can be taught of as essentially Muirhead, since it is clear that $ [ k, 0, k, 0, \ldots, 0, k ] $ majorities $ [ 0, k+1, 0, k+2, \ldots, k+1, 0 ] $.
• Of course, there are other ways that one can do the AM-GM / Jensens.
• Equality holds iff $ x = 1$ (, or $ k = 0$).

Solution 3:

Changing your notation slightly, we have $$ p_{2k}(x) = \frac{2}{2k+1}\left(b_0 + b_1 x + b_1 x^2 + \cdots + b_{2k}x^{2k}+ b_{2k+1} x^{2x+1}\right), $$where $b_n = k$ if $n$ is even and $b_n=k+1$ if $n$ is odd. We will prove the claim by induction on $k$. The base case $n=0$ gives $p_0(x) =0$ and we also have $p_2(x) = 2/3 (x+1)^2$, which is obviously non-negative.

Suppose $p_{2k}(x) \ge 0$, with a minimum at $x=-1$. We have $$ p_{2k+2}(x)- p_{2k}(x) = (x+1)^2\cdot \frac{2}{4k^2-1}\sum_{n=0}^{2k-2} (-1)^n\frac{(n+1)(n+2)}{2} x^n $$Cancelling with the factor of $(x+1)^2$, the sum reduces to $$ S_{2k}(x)=\frac{2 k^2 (x + 1)^2 x^{2k-1} - k (x^2 - 1) x^{2k-1} -x^{2k} + 1}{ x + 1};$$ note that $S_{2k}(0)=1$ and $\lim_{x\to -1} S_{2k}(x) $ exists and is positive by LHR.

We claim the numerator has a unique root at $x=-1$. Indeed, $$ 2 k^2 (x + 1)^2 x^{2k-1} - k (x^2 - 1) x^{2k-1} -x^{2k} + 1$$ $$ = (2k^2-k)x^{2k+1}+(4k^2-1)x^{2k}+(2k^2+k)x^{2k-1}+1, $$ and its derivative is $$ x^{2k-2}\bigg((2k^2-k)(2k+1)x^2 + (4k^2-1)(2k)x + (2k^2+k)(2k-1)\bigg) $$By the quadratic formula, this has a unique double root at $x=-1$. This means the numerator of $S_{2k}$ changes sign at $x=-1$; in particular, the numerator is negative for $x<-1$ and positive for $x>-1$, implying $S_{2k}(x)>0$. In summary, $p_{2k}(x)$ is clearly zero at $x=-1$ and is increasing in $k$ for any fixed $x$, so the inequality holds.