MathDB
x_1^2 +x_2^2+...+x_n^2= 1, [x_1 +x_2 +...+x_n] = m, x_1 +x_2 +...+x_m \ge 1

Source: Romania BMO TST 1992 p4

February 19, 2020
Suminequalitiesalgebra

Problem Statement

Let x1,x2,...,xnx_1,x_2,...,x_n be real numbers with 1x1x2...xn01 \ge x_1 \ge x_2\ge ... \ge x_n \ge 0 and x12+x22+...+xn2=1x_1^2 +x_2^2+...+x_n^2= 1. If [x1+x2+...+xn]=m[x_1 +x_2 +...+x_n] = m, prove that x1+x2+...+xm1x_1 +x_2 +...+x_m \ge 1.