a>b>0,則a>0,b>0且a-b>0.
故依基本不等式得
a+16/[b(a-b)]
≥a+16/[(b+a-b)/2]^2
＝a+64/a^2
＝(a/2)+(a/2)+(64/a^2)
≥3(a/2·a/2·64/a)^(1/3)
＝6×2^(1/3).
∴a/2＝64/a^2且b＝a-b,
即a＝4×2^(1/3),b＝2×2^(1/3)時,
所求最小值為：6×2^(1/3）