證一不等式

using MI

when n=1
LHS=1/2
RHS=sqrt(1/2)

LHS<=RHS

Assume that when n=k, P(k) is true

[1*3*5*...*(2k-1)] / [(2*4*6*...*(2k)] <= 1/sqrt(k+1)

when n=k+1

LHS
=[1*3*5*...*(2k-1)] / [(2*4*6*...*(2n)]*[(2k+1)/(2k+2)]
<= 1/sqrt(k+1) *[(2k+1)/(2k+2)]

Consider
(k+2)(2k+1)^2-(k+1)(2k+2)^2
=(k+2)(4k^2+4k+1)-4(k^3+3k^2+3k+1)
=4k^2+k+8k^2+8k+2-12k^2-12k-4
=-(3k+2)
<0

So
sqrt(k+2)(2k+1)

2008-06-07 16:47:33 補充：
&lt

2008-06-07 21:13:07 補充：
原來我COPY少了東西﹐這個是yahoo的問題

So
sqrt(k+2)(2k+1) < sqrt(k+1)(2k+2)
1/sqrt(k+2) > [1/sqrt(k+1)][(2k+1)/(2k+2)]

THAT IS

[1*3*5*...*(2k-1)] / [(2*4*6*...*(2n)]*[(2k+1)/(2k+2)]
<=1/sqrt(k+1) *[(2k+1)/(2k+2)]
<=1/sqrt(k+2)

when n=k+1 P(n) is true

So by MI, for all positive integer n, P(n) is true

2008-06-07 21:13:40 補充：
the answer by the next one is wrong

Let P(n) be [1*3*5*...*(2n-1)] / [(2*4*6*...*(2n)] &lt;= 1/sqrt(n+1), where n is a positive integer
When n=1
LHS=1/2&lt;(sqrt2)/2=1/(sqrt2)=RHS
∴P(1)is true
Assume that P(k)is true where k is a natural no.
i.e.[1*3*5*...*(2k-1)] / [(2*4*6*...*(2k)] &lt;= 1/sqrt(k+1) for all positive integer k
For n=k+1
LHS=[1*3*5*...*(2k-1)(2k+1)] / [(2*4*6*...*(2n)(2k+2)]
&lt;=(2k+1)/(2k+2)[sqrt(2k+1)] (Induction hypothesis)
=sqrt(2k+1)/(2k+2)
=[sqrt(2k+1)sqrt(2k+2)/(2k+2)][1/sqrt(2k+2)]
=[sqrt(2k+1)/sqrt(2k+2)][1/sqrt(2k+2)]
&lt;=1/sqrt(2k+2) (As sqrt(2k+1)/sqrt(2k+2)&lt;1)
&lt;=1/sqrt(k+2) (As 2k+2 &gt; k+2 &gt; 0 for all k, implies that sqrt(2k+2) &gt; sqrt(k+2) for all k, wich again implies that 1/sqrt(2k+2) &lt;= 1/sqrt(k+2))

2008-06-07 17:02:33 補充：
&lt= less than
&gt= greater than

2008-06-07 17:04:02 補充：
For better understanding, it is better for you to write the steps on a piece of paper