Question

已知a,b為正數(shù),求證:

(1)若+1>,則對(duì)于任何大于1的正數(shù)x,恒有ax+>b成立.

(2)若對(duì)于任何大于1的實(shí)數(shù)x,恒有ax+>b成立,則+1>.

 

Answer 1

見(jiàn)解析

【解析】證明：(1)∵x>1,∴ax+=a(x-1)++1+a≥2+1+a=(+1)2.

∵+1>(b>0),

∴(+1)2>b.

即ax+>b.

(2)∵ax+>b對(duì)于大于1的實(shí)數(shù)x恒成立,

即x>1時(shí),[ax+]min>b,

而ax+=a(x-1)++1+a≥2+1+a=(+1)2,

當(dāng)且僅當(dāng)a(x-1)=,

即x=1+>1時(shí)取等號(hào).

故[ax+]min=(+1)2.

則(+1)2>b,即+1>.