MathDB

China TST 1986 symmetric sum

Source: China TST 1986, problem 7

May 16, 2005

inequalitiesquadraticsinequalities unsolved

Problem Statement

Let

x_i,

p^2 \cdot \frac{n-1}{n}-2q \geq 0

be real numbers with

n \geq 3.

Let

p

and

q

be their symmetric sum of degree

1

and

2

respectively. Prove that: i)

p^2 \cdot \frac{n-1}{n}-2q \geq 0

ii)

\left|x_i - \frac{p}{n}\right| \leq \sqrt{p^2 - \frac{2nq}{n-1}} \cdot \frac{n-1}{n}

for every meaningful

i

.

Back to Problems View on AoPS