An inequality regarding the coefficients of a real polynomial

algebra-precalculusinequalitypolynomials

Let $P=X^n-a_1\cdot X^{n-1}+a_2\cdot X^{n-2}-\cdots+(-1)^n\cdot a_n$ be a polynomial with real coefficients whose roots are all real and in $[0,1]$. Prove the inequality
$$a_k-a_{k+1}+a_{k+2}-\cdots+(-1)^{n-k}\cdot a_n\ge 0, \forall k, 1\le k\le n.$$
I thought about using Vieta's formulas, but this alone doesn't lead anywhere. I think that this inequality should be proveable by mathematical induction, but I think that we should somehow go from $n$ to $1$ since the case $k=n$ is the easiest (we just need to show that $a_n\ge 0$ and this is obvious since $a_n=x_1\cdot x_2 \cdot\ldots\cdot x_n\ge 0$, where $x_i, i=\overline{1,n}$ are $P$'s roots). Anyway, I couldn't make any further progress.

Best Answer

With $a_0 := 1$ we have $$ P(x) = \sum_{k=0}^n (-1)^k a_k x^{n-k} \, . $$ The idea is to define a polynomial whose roots are all negative or zero, because that translates more easily to a condition on its coefficients. So we define $$ Q(y) = \sum_{k=0}^n (-1)^k a_k (y-1)^k y^{n-k} \, . $$ $Q$ is a polynomial of degree at most $n$, $Q(1) = a_0 = 1$, and $$ Q(y) = (y-1)^n P\left( \frac{y}{y-1}\right) $$ for $y \ne 1$. If $Q(y) = 0$ then $\frac{y}{y-1}$ is real and in the range $[0, 1]$, and that implies $y \le 0$, so:

  1. All roots of $Q$ are real and negative or zero.

Also $P(x) > 0$ for $x > 1$ implies $Q(y) > 0$ for $y > 1$, therefore:

  1. The leading coefficient of $Q$ is positive.

It follows from 1 and 2 that $Q(0) \ge 0$. Repeated application of Rolle's theorem shows that $Q', Q'', \ldots$ also have the properties 1 and 2, so we have

  1. $Q(y) = \sum_{j=0}^n b_j y^{n-j}$ with nonnegative coefficients $b_0, \ldots b_n$.

Now we want to express the $a_k$ in terms of the $b_j$. Since the mapping $t \mapsto \frac{t}{t-1}$ is self-inverse, it is not difficult to see that $$ P(x)=(x-1)^n Q\left( \frac{x}{x-1}\right) $$ for $x \ne 1$, and therefore $$ \begin{align} P(x) &= \sum_{k=0}^n (-1)^k x^{n-k} \sum_{j=k}^n \binom j k b_j \\ &= \sum_{j=0}^n b_j x^{n-j} \sum_{k=0}^j \binom j k (-1)^{k} x^{j-k} \\ &= \sum_{k=0}^n (-1)^k x^{n-k} \sum_{j=k}^n \binom j k b_j \, . \end{align} $$ Comparing this with the original formula for $P$ gives

  1. $a_k = \sum_{j=k}^n \binom j k b_j$ for $k=0, \ldots, n$.

Finally we can compute $$ \sum_{k=i}^n (-1)^{k-i}a_k = \sum_{k=i}^n (-1)^{k-i}\sum_{j=k}^n \binom j k b_j = \sum_{j=i}^n b_j \sum_{k=i}^j \binom j k (-1)^{k-i} \, , $$ and since all $b_j$ are nonnegative it suffices to show that $$ c_{i, j} = \sum_{k=i}^j \binom j k (-1)^{k-i} \ge 0 $$ for $0 \le i \le j \le n$. For $j=0$ is $c_{0, 0} = 1$. For $j > 0$ we substitute $k=j-l$ and use the formula from Truncated alternating binomial sum: $$ c_{i, j} = \sum_{l=0}^{j-i} \binom{j}{j-l}(-1)^{j-l-i} = (-1)^{j-i} \sum_{l=0}^{j-i} \binom j l (-1)^l = \binom{j-1}{j-i} \ge 0 \, , $$ and that finishes the proof.

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