$n$ is a real number and $n\ge1$.Prove that \[ \frac {1}{2^{\frac {1}{n} }-1} \ge n\]
with equality if and only if $n=1$.
A JUNIOR TYPE PROBLEM.NOT FOR OTHERS.
Inequality !!! [self-made]
- Phlembac Adib Hasan
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Re: Inequality !!! [self-made]
(n+1)^n (1)
=n^n+C(n,1)xn^(n-1)+C(n,20)xn^(n-2)+.....+1
=n^n+nxn^(n-1)+.....+1
=n^n+n^n+C(n,2)xn^(n-2)+.....+1 (2)
From (1) and (2),we can say,
(n+1)^n>n^n+n^n [For n>1]
or,(n+1)^n>2xn^n
or,((n+1)^n)/(n^n)>2
or,((n+1)/n)^n>2
or,(n+1)/n>2^(1/n)
or,1+(1/n)>2^(1/n)
or,1/n>2^(1/n)-1
or, 1/( 2^(1/n)-1)>n
Equality holds only when n=1 and this can be veryfied. [Proved]
=n^n+C(n,1)xn^(n-1)+C(n,20)xn^(n-2)+.....+1
=n^n+nxn^(n-1)+.....+1
=n^n+n^n+C(n,2)xn^(n-2)+.....+1 (2)
From (1) and (2),we can say,
(n+1)^n>n^n+n^n [For n>1]
or,(n+1)^n>2xn^n
or,((n+1)^n)/(n^n)>2
or,((n+1)/n)^n>2
or,(n+1)/n>2^(1/n)
or,1+(1/n)>2^(1/n)
or,1/n>2^(1/n)-1
or, 1/( 2^(1/n)-1)>n
Equality holds only when n=1 and this can be veryfied. [Proved]
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: Inequality !!! [self-made]
Correct proof, but long.Notice that we can say the statement $( \frac {n+1} {n})^n >2$ from the property of e.The following part is similar to me.SANZEED wrote:(n+1)^n (1)
=n^n+C(n,1)xn^(n-1)+C(n,20)xn^(n-2)+.....+1
=n^n+nxn^(n-1)+.....+1
=n^n+n^n+C(n,2)xn^(n-2)+.....+1 (2)
From (1) and (2),we can say,
(n+1)^n>n^n+n^n [For n>1]
or,(n+1)^n>2xn^n
or,((n+1)^n)/(n^n)>2
or,((n+1)/n)^n>2
or,(n+1)/n>2^(1/n)
or,1+(1/n)>2^(1/n)
or,1/n>2^(1/n)-1
or, 1/( 2^(1/n)-1)>n
Equality holds only when n=1 and this can be veryfied. [Proved]
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