f6_prure mathe _9

let p(n) be the proposition n^(1/n) <= (n+2(n^(1/2))-2)/n.
when n=1

LHS=(1)^(1/1)
=1
RHS=(1+2(1^(1/2))-2)/1
=1

So p(1) is true.

Assume p(k) is ture,
i.e. k^(1/k) <= (k+2(k^(1/2))-2)/k

when n=k+1
LHS
=(k+1)^(1/(k+1))
<=k^(1/k)
<=(k+2k^(1/2)-2)/k
<=(k+1+2(k+1)^(1/2)-2)/(k+1)
=RHS

so p(k+1) is true.

According to MI,when n is positive integer,p(n) is true
Prove that (k+2(k^(1/2))-2)/k<=(k+1+2(k+1)^(1/2)-2)/(k+1)
is equivalent to (k+1)(k+2(k^(1/2))-2)<=k(k+1+2(k+1)^(1/2)-2)
That is
k^2+2k^(3/2)-2k+k+2k^(1/2)-2<=k^2+k+2k(k+1)^(1/2)-2
<=>k^(3/2)-k+k^(1/2)<=k(k+1)^(1/2)
<=>k-k^(1/2)+1<=[k(k+1)]^(1/2)
<=>k+1<=k^(1/2)[1+(k+1)^(1/2)]...(*)
But since k+1<=k^(1/2)[1+k^(1/2)] for n>1
So (*) should be true

prove
=(k+1)^(1/(k+1))

<=k^(1/k)

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