放缩法的不等式证明技巧
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放缩法的不等式证明技巧,我举几例子给你看看:
1、3/2*5/4*7/6*9/8*...2n+1/2n>√(n+1)
3/2*5/4*7/6*9/8*...2n+1/2n)²
=(3/2*5/4*7/6*9/8*...2n+1/2n)(3/2*5/4*7/6*9/8*...2n+1/2n)
>(3/2*5/4*7/6*9/8*...2n+1/2n)(4/3*6/5*8/7*10/9...(2n+2)(2n+1))
=(2n+2)/2=n+1
∴ 3/2*5/4*7/6*9/8*...2n+1/2n>√(n+1)
2、证明|sin(x-y)|≤|sin(x-z)|+|sin(z-y)|
∵|cosx|≤1
|sin(x-y)|=|sin(x-y-z+z)|=|sin[(x-z)-(y-z)]|
=|sin[(x-z)cos(y-z)-cos(x-z)sin(y-z)|
=|sin[(x-z)cos(y-z)+cos(x-z)sin(z-y)|
≤|sin[(x-z)cos(y-z)|+|cos(x-z)sin(z-y)|
≤|sin[(x-z)|+|sin(z-y)|
1、3/2*5/4*7/6*9/8*...2n+1/2n>√(n+1)
3/2*5/4*7/6*9/8*...2n+1/2n)²
=(3/2*5/4*7/6*9/8*...2n+1/2n)(3/2*5/4*7/6*9/8*...2n+1/2n)
>(3/2*5/4*7/6*9/8*...2n+1/2n)(4/3*6/5*8/7*10/9...(2n+2)(2n+1))
=(2n+2)/2=n+1
∴ 3/2*5/4*7/6*9/8*...2n+1/2n>√(n+1)
2、证明|sin(x-y)|≤|sin(x-z)|+|sin(z-y)|
∵|cosx|≤1
|sin(x-y)|=|sin(x-y-z+z)|=|sin[(x-z)-(y-z)]|
=|sin[(x-z)cos(y-z)-cos(x-z)sin(y-z)|
=|sin[(x-z)cos(y-z)+cos(x-z)sin(z-y)|
≤|sin[(x-z)cos(y-z)|+|cos(x-z)sin(z-y)|
≤|sin[(x-z)|+|sin(z-y)|
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