Inequality proof need help understanding solution

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Proof that $(sqrtn)^sqrtn+1>(sqrtn+1)^sqrtn.$
For values $n= 7,8,9....$

The solution is given as follows. Could anyone help to explain this? (Especially the last step "Consequently...")

Also, the question had mentioned (but had not stated) that the proof by calculus is possible. Could anyone also show this?

inequality contest-math proof-explanation

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edited Jul 14 at 14:22

amWhy

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asked Jul 14 at 12:58

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up vote
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Proof that $(sqrtn)^sqrtn+1>(sqrtn+1)^sqrtn.$
For values $n= 7,8,9....$

The solution is given as follows. Could anyone help to explain this? (Especially the last step "Consequently...")

Also, the question had mentioned (but had not stated) that the proof by calculus is possible. Could anyone also show this?

inequality contest-math proof-explanation

share|cite|improve this question

edited Jul 14 at 14:22

amWhy

189k25219431

asked Jul 14 at 12:58

299792458

185

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Proof that $(sqrtn)^sqrtn+1>(sqrtn+1)^sqrtn.$
For values $n= 7,8,9....$

The solution is given as follows. Could anyone help to explain this? (Especially the last step "Consequently...")

Also, the question had mentioned (but had not stated) that the proof by calculus is possible. Could anyone also show this?

inequality contest-math proof-explanation

share|cite|improve this question

edited Jul 14 at 14:22

amWhy

189k25219431

asked Jul 14 at 12:58

299792458

185

Proof that $(sqrtn)^sqrtn+1>(sqrtn+1)^sqrtn.$
For values $n= 7,8,9....$

The solution is given as follows. Could anyone help to explain this? (Especially the last step "Consequently...")

Also, the question had mentioned (but had not stated) that the proof by calculus is possible. Could anyone also show this?

inequality contest-math proof-explanation

share|cite|improve this question

edited Jul 14 at 14:22

amWhy

189k25219431

asked Jul 14 at 12:58

299792458

185

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 14:22

amWhy

189k25219431

edited Jul 14 at 14:22

amWhy

189k25219431

edited Jul 14 at 14:22

amWhy

189k25219431

189k25219431

asked Jul 14 at 12:58

299792458

185

asked Jul 14 at 12:58

299792458

185

asked Jul 14 at 12:58

299792458

185

185

add a comment | 

add a comment | 

1 Answer
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Consider that
$$sqrt n^sqrt n+1 >sqrt n+1^sqrt nimplies sqrt n+1log(sqrt n )>sqrt nlog(sqrt n+1 )$$ that is to say
$$fraclog(sqrt n ) sqrt n >fraclog(sqrt n+1 )sqrt n+1$$

In the real domain, consider the function
$$f(x)=fraclog(sqrt x ) sqrt x implies f'(x)=-fraclog (x)-24 x^3/2$$ The first derivative cancels at $x=e^2$ and the second deriative test shows that this is a maximum.

Now $e^2 approx 7.38$ could help.

share|cite|improve this answer

edited Jul 14 at 14:22

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Consider that
$$sqrt n^sqrt n+1 >sqrt n+1^sqrt nimplies sqrt n+1log(sqrt n )>sqrt nlog(sqrt n+1 )$$ that is to say
$$fraclog(sqrt n ) sqrt n >fraclog(sqrt n+1 )sqrt n+1$$

In the real domain, consider the function
$$f(x)=fraclog(sqrt x ) sqrt x implies f'(x)=-fraclog (x)-24 x^3/2$$ The first derivative cancels at $x=e^2$ and the second deriative test shows that this is a maximum.

Now $e^2 approx 7.38$ could help.

share|cite|improve this answer

edited Jul 14 at 14:22

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

add a comment | 

up vote
0
down vote

Consider that
$$sqrt n^sqrt n+1 >sqrt n+1^sqrt nimplies sqrt n+1log(sqrt n )>sqrt nlog(sqrt n+1 )$$ that is to say
$$fraclog(sqrt n ) sqrt n >fraclog(sqrt n+1 )sqrt n+1$$

In the real domain, consider the function
$$f(x)=fraclog(sqrt x ) sqrt x implies f'(x)=-fraclog (x)-24 x^3/2$$ The first derivative cancels at $x=e^2$ and the second deriative test shows that this is a maximum.

Now $e^2 approx 7.38$ could help.

share|cite|improve this answer

edited Jul 14 at 14:22

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

add a comment | 

up vote
0
down vote

up vote
0
down vote

Consider that
$$sqrt n^sqrt n+1 >sqrt n+1^sqrt nimplies sqrt n+1log(sqrt n )>sqrt nlog(sqrt n+1 )$$ that is to say
$$fraclog(sqrt n ) sqrt n >fraclog(sqrt n+1 )sqrt n+1$$

In the real domain, consider the function
$$f(x)=fraclog(sqrt x ) sqrt x implies f'(x)=-fraclog (x)-24 x^3/2$$ The first derivative cancels at $x=e^2$ and the second deriative test shows that this is a maximum.

Now $e^2 approx 7.38$ could help.

share|cite|improve this answer

edited Jul 14 at 14:22

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

Consider that
$$sqrt n^sqrt n+1 >sqrt n+1^sqrt nimplies sqrt n+1log(sqrt n )>sqrt nlog(sqrt n+1 )$$ that is to say
$$fraclog(sqrt n ) sqrt n >fraclog(sqrt n+1 )sqrt n+1$$

In the real domain, consider the function
$$f(x)=fraclog(sqrt x ) sqrt x implies f'(x)=-fraclog (x)-24 x^3/2$$ The first derivative cancels at $x=e^2$ and the second deriative test shows that this is a maximum.

Now $e^2 approx 7.38$ could help.

share|cite|improve this answer

edited Jul 14 at 14:22

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 14:22

edited Jul 14 at 14:22

edited Jul 14 at 14:22

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

answered Jul 14 at 14:02

Claude Leibovici

112k1055126

112k1055126

add a comment | 

add a comment | 

 

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