感觉应该求M的最大值
a²+b²≥M√(ab-a-b+1)恒成立
即M≤(a²+b²)/√(ab-a-b+1)恒成立
需M≤ [(a²+b²)/√(ab-a-b+1)]min
∵a,b>1,∴a-1>0,b-1>0
∵ a²+b²
=(a-1)²+2(a-1)+1+(b-1)²+2(b-1)+1
=(a-1)²+(b-1)²+2(a-1)+2(b-1)+2
∵(a-1)²+(b-1)²≥2(a-1)(b-1)
2(a-1)+2(b-1)≥ 2√[4(a-1)(b-1)]=4√[(a-1)(b-1)]
∴ a²+b²≥2(a-1)(b-1) +4√[(a-1)(b-1)]+2
∵√(ab-a-b+1)=√[(a-1)(b-1)]
∴(a²+b²)/√(ab-a-b+1)≥2√[(a-1)(b-1)+2/√[(a-1)(b-1)]+4
∵ 2√[(a-1)(b-1)+2/√[(a-1)(b-1)]≥2×2=4
∴(a²+b²)/√(ab-a-b+1)≥4+4=8
即[(a²+b²)/√(ab-a-b+1)]min =8
∴M≤8
即M的最大值为8
已知a,b>1,a2+b2≥M根号(ab-a-b+1)恒成立,求M的最小值
已知a,b>1,a2+b2≥M根号(ab-a-b+1)恒成立,求M的最小值
数学人气:763 ℃时间:2020-03-24 08:23:26
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