... Now that16x + 1 = (x2− y2)2= (x −y)2(x + y)2≥ x2.From this we obtain the inequality, x2− 16x − 1 < 0. Solving this inequality gives x ∈{0, 1, ··· ,16}. In addition, 16x + 1 is ... y) = |x − y| + |x3− y3|when x, y vary satisfying the restriction x2+ y2= 1.Rewriting this asf(x, y) = |x − y|(1 + x2+ xy + y2) = |x − y|(2 + xy).from which we square to arrive ... xy33=533.Hence,f(x, y) ≤53.53.Equality occurs whenxy = −13, x2+ y2= 1.This simultaneous equations are equivalent toxy = −13, x + y =1√3.5Hanoi Open Mathematical...