Let $\alpha_1, \alpha_2, \dots, \alpha_n$, and $\beta_1, \beta_2, \ldots, \beta_n$, where $n \geq 4$, be 2 sets of real numbers such that \[\sum_{i=1}^{n} \alpha_i^2 < 1 \qquad \text{and} \qquad \sum_{i=1}^{n} \beta_i^2 < 1.\]Define \begin{align*} A^2 &= 1 - \sum_{i=1}^{n} \alpha_i^2,\\ B^2 &= 1 - \sum_{i=1}^{n} \beta_i^2,\\ W &= \frac{1}{2} (1 - \sum_{i=1}^{n} \alpha_i \beta_i)^2. \end{align*}Find all real numbers $\lambda$ such that the polynomial \[x^n + \lambda (x^{n-1} + \cdots + x^3 + Wx^2 + ABx + 1) = 0,\]only has real roots.
Problem
Source: China TST 1996, problem 5
Tags: inequalities, algebra, Sums and Products, Real Roots, China
who
19.06.2006 13:47
,motivated from bulgaria final round pro. idea is common to get equalities in classical inequalities,try which. :
brdkc43
13.08.2016 02:22
My Solution, Please Check this solution! We will show for infinitely many solution for $\lambda$ is provide the conditions. Let's begin. From equations in the problem we have $$A^2+B^2+\sum_{i=1}^{n}\alpha_i^2+\sum_{i=1}^{n}\beta_i^2=2\ge 2AB+\sum_{i=1}^{n}\alpha_i^2+\sum_{i=1}^{n}\beta_i^2\ge 2AB+2\left( \sum_{i=1}^{n}\alpha_i\beta_i\right)$$ $$=2\left( AB+\sum_{i=1}^{n}\alpha_i\beta_i\right)\le 2 \Longleftrightarrow AB+\sum_{i=1}^{n}\alpha_i\beta_i\le 1 = \sqrt{\left( 1 - \sum_{i=1}^{n}\alpha_i^2\right)\left( 1 - \sum_{i=1}^{n}\beta_i^2\right)}+\sum_{i=1}^{n}\alpha_i\beta_i\le 1$$ $$1+\left( \sum_{i=1}^{n}\alpha_i^2+\sum_{i=1}^{n}\beta_i^2\right)+ \sum_{i=1}^{n}\sum_{i=1}^{n}\alpha_i^2\beta_i^2 \le 1+ \left( \sum_{i=1}^{n}\alpha_i\beta_i\right)^2-2\left( \sum_{i\not=j}^{n}\alpha_i \beta_j\right)$$ İts easy to see, $$\left(\sum_{i=1}^{n}\alpha_i^2\right)\left(\sum_{i=1}^{n}\beta_i^2\right)+2\sum_{i\not=j}\alpha_i\beta_j\le \left( \sum_{i=1}^{n}a_ib_i \right)^2 +\sum_{i=1}^{n}\alpha^2+\sum_{i=1}^{n}\beta_i^2 \Rightarrow \left( \sum_{i=1}^{n}\alpha_i\right)\left( \sum_{i=1}^{n}\beta_i\right)\le \sum_{i=1}^{n}\alpha^2+\sum_{i=1}^{n}\beta_i^2<2$$Cause we have $$\left( \sum_{i=1}^{n}a_ib_i \right)^2\ge 2\sum_{i\not=j}\alpha_i\beta_j$$ For simplicity we put, $n=2k$ and $\lambda = -n$, let $x_i$ be root of equation. Put prime factors of $n$ to, $x_1,x_2,\dots$ then if number of prime factors of $n$ smaller then number of roots, we put ${-1,1}$ to this roots and we have all real roots. And We ar e done!