Two real sequence {xn} and {yn} satisfies following recurrence formula;
x0=1 , y0=2007
xn+1=xn−(xnyn+xn+1yn+1−2)(yn+yn+1) ,
yn+1=yn−(xnyn+xn+1yn+1−2)(xn+xn+1)
Then show that for all nonnegative integer n , xn2≤2007 .
Dataset text AI-MO AoPS. Quoted with source.
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