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4

Part of 1992 Romania Team Selection Test

Problems(2)

1992−th term of the ordered set A of all ordered sequences (a_1,a_2,...,a_{11})

Source: Romania IMO TST 1992 p4

2/19/2020
Let AA be the set of all ordered sequences (a1,a2,...,a11)(a_1,a_2,...,a_{11}) of zeros and ones. The elements of AA are ordered as follows: The first element is (0,0,...,0)(0,0,...,0), and the n+1n + 1−th is obtained from the nn−th by changing the first component from the right such that the newly obtained sequence was not obtained before. Find the 19921992−th term of the ordered set AA
Sequencecombinatoricsalgebraset
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

2/19/2020
Let x1,x2,...,xnx_1,x_2,...,x_n be real numbers with 1≥x1≥x2≥...≥xn≥01 \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+...+xm≥1x_1 +x_2 +...+x_m \ge 1.
Suminequalitiesalgebra