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Berkeley Math Circle: Monthly Contest 5

Algebra Difficulty 4.9 AIME Prove it United States

Problem:
Let a1,a2,…a_{1}, a_{2}, \ldots be an infinite sequence of positive real numbers which satisfies
an+1≥an2+15 a_{n+1} \geq a_{n}^{2}+\frac{1}{5}
for every positive integer nn. Prove that an+5≥an−5\sqrt{a_{n+5}} \geq a_{n-5} for each positive integer nn.

Solution

Solution:
From the given we can deduce that
an+1≥an2+14−120≥an−120. a_{n+1} \geq a_{n}^{2}+\frac{1}{4}-\frac{1}{20} \geq a_{n}-\frac{1}{20}.
Thus for any nn we have
an+5≥an+1−4⋅120=an+1−15=an2 a_{n+5} \geq a_{n+1}-4 \cdot \frac{1}{20}=a_{n+1}-\frac{1}{5}=a_{n}^{2}
Thus an+5≥an≥an−5\sqrt{a_{n+5}} \geq a_{n} \geq a_{n-5} follows.

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