设函数f(x)=(x+1)ln(x+1)-ax在x=0处取得极值,证明对任意的正整数n,不等式nlnn>=(n-1)ln(n+1)都成立
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f(x)= (x+1)ln(x+1)-nx
f'(x) = 1-n ≤ 0
f is decreasing function
f(0) = 0
f(n-1) = nlnn- n(n-1)
f(n) = (n+1)ln(n+1) - (n+1)n
f(n)≤ f(n-1)
=>(n+1)ln(n+1) - (n+1)n≤nlnn- n(n-1)
=>nlnn≥(n+1)ln(n+1) - (n+1)n + n(n-1)
= (n+1)ln(n+1) -2n
= (n-1)ln(n+1) - 2[n -ln(n+1)]
≥(n-1)ln(n+1) ( n > ln(n+1) )
f'(x) = 1-n ≤ 0
f is decreasing function
f(0) = 0
f(n-1) = nlnn- n(n-1)
f(n) = (n+1)ln(n+1) - (n+1)n
f(n)≤ f(n-1)
=>(n+1)ln(n+1) - (n+1)n≤nlnn- n(n-1)
=>nlnn≥(n+1)ln(n+1) - (n+1)n + n(n-1)
= (n+1)ln(n+1) -2n
= (n-1)ln(n+1) - 2[n -ln(n+1)]
≥(n-1)ln(n+1) ( n > ln(n+1) )
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