問題詳情:
設x≥1,y≥1,*:x+y++xy;
【回答】
*:(1)由於x≥1,y≥1,
所以x+y++xy⇔xy(x+y)+1≤y+x+(xy)2,
此式的右邊減去左邊得
y+x+(xy)2-[xy(x+y)+1]
=[(xy)2-1]-[xy(x+y)-(x+y)]
=(xy+1)(xy-1)-(x+y)(xy-1)
=(xy-1)(xy-x-y+1)
=(xy-1)(x-1)·(y-1).
因為x≥1,y≥1,所以(xy-1)(x-1)(y-1)≥0.
故所*不等式成立.
知識點:不等式
題型:解答題