Let $P(x)$ be a monic polynomial of degree $n$ with nonnegative coefficients and the free term equal to $1$. Prove that if all the roots of $P(x)$ are real, then $P(x) \geq (x+1)^{n}$ holds for every $x \geq 0$.
Problem
Source: Croatian NMC 2005, 4 th Grade
Tags: algebra, polynomial, limit, inequalities, algebra unsolved
07.05.2007 16:42
From hypothesis we have $P(x)=(x+x_{1})(x+x_{2})...(x+x_{n})$, here $x_{i}\geq 0\; \forall i$ and $x_{1}x_{2}...x_{n}=1$. If $x=\frac{p}{q}\in\mathbb{Q}\cap [0,\infty)$ then by AM-GM we have $P(\frac{p}{q})=\frac{1}{q^{n}}\cdot \prod_{i=1}^{n}(p+qx_{i})\geq \frac{1}{q^{n}}\cdot \prod_{i=1}^{n}(p+q)\cdot \sqrt[p+q]{x_{i}^{q}}=(\frac{p}{q}+1)^{n}$. Now by : for all $x\in [0;\infty)$ there exists sequence of nonnegagive rationals $(q_{n})$ such that $\lim_{n\to\infty}q_{n}=x$ we're done!
14.06.2008 01:53
we have that $ P(x)\geq 1$ for all $ x\geq 0$, so all the roots of $ P$ are negative. Let $ P(x)=\prod_{i=1}^n(x+\alpha_i)$, with $ \alpha_i>0$ for all $ i$. from the hypothesis we have that $ \prod_{i=1}^n\alpha_i=1$, then, from holder's inequality we have that $ P(x)\geq \left(x+\sqrt[n]{\prod_{i=1}^n\alpha_i}\right)^n=(x+1)^n$, as we wanted to show.
15.06.2008 13:40
A question for the two solutions above,How can assert the roots of $ P(x)$ is all negative?In the problem it just state the roots is real and the coefficients is nonnegative,doesn't state that the strong condition:all the roots is negative.
15.06.2008 14:10
zhaobin wrote: A question for the two solutions above,How can assert the roots of $ P(x)$ is all negative?In the problem it just state the roots is real and the coefficients is nonnegative,doesn't state that the strong condition:all the roots is negative. Doesn't non-negativity of coefficients imply positivity of roots? If all coefficients are non-negative and at least one of them is positive(in our case the free term is equal to $ 1$),then polynomial takes only positive values on positive line...
15.06.2008 22:21
suppose there is a positive root, this is $ (x-\alpha)$ divides $ P(x)$ for some $ \alpha\geq 0$. we have that $ 0=P(\alpha)=\sum_{k=0}^na_k\alpha^k\geq a_0=1$ which is absurd... is this better?
18.06.2008 11:32
Yes,Thank you,campos.It will be better.
08.07.2008 08:55
N.T.TUAN wrote: From hypothesis we have $ P(x) = (x + x_{1})(x + x_{2})...(x + x_{n})$, here $ x_{i}\geq 0\; \forall i$ and $ x_{1}x_{2}...x_{n} = 1$. If $ x = \frac {p}{q}\in\mathbb{Q}\cap [0,\infty)$ then by AM-GM we have $ P(\frac {p}{q}) = \frac {1}{q^{n}}\cdot \prod_{i = 1}^{n}(p + qx_{i})\geq \frac {1}{q^{n}}\cdot \prod_{i = 1}^{n}(p + q)\cdot \sqrt [p + q]{x_{i}^{q}} = (\frac {p}{q} + 1)^{n}$. Now by : for all $ x\in [0;\infty)$ there exists sequence of nonnegagive rationals $ (q_{n})$ such that $ \lim_{n\to\infty}q_{n} = x$ we're done! I don't see how $ P(\frac {p}{q}) = \frac {1}{q^{n}}\cdot \prod_{i = 1}^{n}(p + qx_{i})\geq \frac {1}{q^{n}}\cdot \prod_{i = 1}^{n}(p + q)\cdot \sqrt [p + q]{x_{i}^{q}}$
08.07.2008 08:58
$ p + qx_i = \underbrace{1 + 1 + ...}_{p \text{ times}} + \underbrace{x_i + x_i + ...}_{q \text{ times}} \ge (p + q) \sqrt[p+q]{1^p x_i^q}$. This is generalized by weighted AM-GM.
27.02.2019 15:52
this problem used in Hungary_1983
06.05.2020 00:31
Here is a stronger proof - lower bounding each coefficient. Let $P(x)=\sum_{k=0}^n \alpha_k x^k$ with $\alpha_0=\alpha_n=1$, $\alpha_i\geqslant 0$. In particular, roots $r_1,\dots,r_n$ are negative, let $\theta_i=-\alpha_i$. Then $\prod_i \theta_i=1$. Note that using AM-GM, $$ \alpha_{n-1}=-\sum_{1\leqslant i\leqslant n}r_i\geqslant n, \quad \alpha_{n-2}=\sum_{i<j}r_ir_j\geqslant \binom{n}{2}, $$and continuing in this manner, $$ \alpha_k\geqslant \binom{n}{n-k}. $$Thus on $[0,\infty)$, it holds that $$ P(x)\geqslant \sum_{0\leqslant k\leqslant n}\binom{n}{k}x^k=(x+1)^n. $$