Let real coefficient polynomial $f(x) = x^n + a_1 \cdot x^{n-1} + \ldots + a_n$ has real roots $b_1, b_2, \ldots, b_n$, $n \geq 2,$ prove that $\forall x \geq max\{b_1, b_2, \ldots, b_n\}$, we have \[f(x+1) \geq \frac{2 \cdot n^2}{\frac{1}{x-b_1} + \frac{1}{x-b_2} + \ldots + \frac{1}{x-b_n}}.\]
Problem
Source: China TST 1991, problem 1
Tags: algebra, polynomial, inequalities, function, algebra unsolved
harazi
27.06.2005 17:17
Put $x-b_i=a_i$. We must prove that for positive numbers $a_i$ we have $ (1+a_1)...(1+a_n)(1/a_1+...+1/a_n)\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $n(1+G)^n/G$ where $G$ is the geometric mean of $a_i$. A study of the function $\frac{(1+x)^n}{x}$ shows that it has minimal value in $\frac{1}{n-1}$ and this one is $ (n-1)(n/(n-1))^n$ obviously greater than $2n$.
math10
31.10.2009 12:08
harazi wrote: Put $ x - b_i = a_i$. We must prove that for positive numbers $ a_i$ we have $ (1 + a_1)...(1 + a_n)(1/a_1 + ... + 1/a_n)\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $ n(1 + G)^n/G$ where $ G$ is the geometric mean of $ a_i$. A study of the function $ \frac {(1 + x)^n}{x}$ shows that it has minimal value in $ \frac {1}{n - 1}$ and this one is $ (n - 1)(n/(n - 1))^n$ obviously greater than $ 2n$. Can anyone post other solution?(This problem is not difficult but I can not solve this problem by other)
Vincent Gilbert
31.10.2009 16:15
Well , you can prove the last ineq : $ \frac{(1+x)^n}{x} \ge 2n$ like this $ \frac{(1+x)^n}{x} \ge \frac{1}{x}+n+\frac{ n(n-1)}{2x} \ge n+\sqrt{ 2n(n-1)} \ge 2n$
Mathias_DK
01.11.2009 19:10
math10 wrote: harazi wrote: Put $ x - b_i = a_i$. We must prove that for positive numbers $ a_i$ we have $ (1 + a_1)...(1 + a_n)(1/a_1 + ... + 1/a_n)\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $ n(1 + G)^n/G$ where $ G$ is the geometric mean of $ a_i$. A study of the function $ \frac {(1 + x)^n}{x}$ shows that it has minimal value in $ \frac {1}{n - 1}$ and this one is $ (n - 1)(n/(n - 1))^n$ obviously greater than $ 2n$. Can anyone post other solution?(This problem is not difficult but I can not solve this problem by other) $ x-b_i = c_i \Rightarrow (c_1+1)...(c_n+1)\left ( \frac{1}{c_1} + ... + \frac{1}{c_n} \right ) \ge 2n^2$ $ \frac{1}{c_1} + ... + \frac{1}{c_n} \ge n\sqrt[n]{\frac{1}{c_1c_2...c_n}}$ $ c_i+1 = c_i+(n-1)\frac{1}{n-1} \ge n\sqrt[n]{c_i\frac{1}{(n-1)^{n-1}}}$ So $ LHS \ge n^{n+1}\frac{1}{(n-1)^{n-1}} \ge 2n^2 \iff$ $ \left ( \frac{n}{n-1} \right )^{n-1} \ge 2 \iff$ $ \left ( 1 + \frac{1}{m} \right )^m \ge 2, m = n-1 \ge 1$ But $ \left ( 1 + \frac{1}{m} \right )^m = \sum_{i=0}^m \binom{m}{i} m^{-i} \ge \binom{m}{0}+\binom{m}{1}m^{-1} = 2$, and we are done.