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Algebra Difficulty 4.6 AIME Prove it Croatia

Let nn be a positive integer. Prove that for all x1,x2,…,xn∈[0,1]x_1, x_2, \dots, x_n \in [0, 1] the following holds:
(x1+x2+⋯+xn+1)2≥4(x12+x22+⋯+xn2).(ASU 1979) (x_1 + x_2 + \dots + x_n + 1)^2 \ge 4(x_1^2 + x_2^2 + \dots + x_n^2). \quad (\text{ASU 1979})

Solution

Since xi∈[0,1]x_i \in [0, 1], we conclude that xi≥xi2x_i \ge x_i^2, i.e.
4(x1+x2+⋯+xn)≥4(x12+x22+⋯+xn2). 4(x_1 + x_2 + \dots + x_n) \ge 4(x_1^2 + x_2^2 + \dots + x_n^2).
Denote S=x1+x2+⋯+xnS = x_1 + x_2 + \dots + x_n. Notice that, because of the previous bound, it suffices to show that
(S+1)2≥4S. (S + 1)^2 \ge 4S.
This inequality obviously holds since it is equivalent to (S−1)2≥0(S - 1)^2 \ge 0.

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