MathLabs

Problem 1

Determine all sequences of real numbers (a1,a2,…,a1995)(a_1,a_2,\ldots,a_{1995}) satisfying 2an−(n−1)≥an+1−(n−1)2\sqrt{a_n-(n-1)}\ge a_{n+1}-(n-1) for n=1,2,…,1994n=1,2,\ldots,1994, and 2a1995−1994≥a1+12\sqrt{a_{1995}-1994}\ge a_1+1.
Step 3 of 4: Complete the square and force equality
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0≥∑n=11995(xn−2xn+1)=∑n=11995(xn−1)2≥00\ge\sum_{n=1}^{1995}(x_n-2\sqrt{x_n}+1)=\sum_{n=1}^{1995}(\sqrt{x_n}-1)^2\ge0
Detailed analysis

Rearranging the summed inequality gives the leftmost bound. Since xn−2xn+1=(xn−1)2≥0x_n-2\sqrt{x_n}+1=(\sqrt{x_n}-1)^2\ge0, both inequalities can hold only when every square is zero. Therefore xn=1x_n=1 for every nn.