若x≥1,y≥1,则（x-1)(y-1)=xy-x-y+1≥0,
∴xy≥x+y-1.于是
若实数a1,a2,...,a8均不小于1 ,
则a1a2a3a4,a5a6a7a8不小于1,
∴a1a2…a8≥a1a2a3a4+a5a6a7a8-1,
同理a1a2a3a4≥a1a2+a3a4-1,
a5a6a7a8≥a5a6+a7a8-1,
则a1a2…a8≥a1a2a3a4+a5a6a7a8-1≥a1a2+a3a4-1+a5a6+a7a8-1-1≥a1a2+a3a4+a5a6+a7a8-3
同理：
a1a2≥a1+a2-1,
a3a4≥a3+a4-1,
a5a6≥a5+a6-1,
a7a8≥a7+a8-1,
∴a1a2…a8≥a1+a2+a3+a4+a5+a6+a7+a8-7=13,
这与“a1a2...a8