a＞b＞0,即 a＞0,a-b＞0.于是 b(a-b)≤[(b+a-b)/2]^2 = a^2/4 (当且仅当 b = a-b = a/2 时取等号),故 16/b(a-b)≥16/(a^2/4 )= 64/a^2,则 a^2 + 16/b(a-b)≥a^2 + 64/a^2 ≥ 2* 根号下(a^2*64/a^2)= 16 (当且仅当 a^...