inequality with power function












0












$begingroup$



Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$




Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)






inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:47










  • $begingroup$
    It is the way around, see math.stackexchange.com/questions/1990936/….
    $endgroup$
    – Martin R
    Nov 27 '18 at 20:00










  • $begingroup$
    alright thanks for your help
    $endgroup$
    – StefanWK
    Nov 27 '18 at 20:16
















0












$begingroup$



Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$




Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)






inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:47










  • $begingroup$
    It is the way around, see math.stackexchange.com/questions/1990936/….
    $endgroup$
    – Martin R
    Nov 27 '18 at 20:00










  • $begingroup$
    alright thanks for your help
    $endgroup$
    – StefanWK
    Nov 27 '18 at 20:16














0












0








0





$begingroup$



Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$




Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)






inequality







share|cite|improve this question











$endgroup$





Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$




Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 19:47









user376343

3,4033826




3,4033826










asked Nov 27 '18 at 19:29









StefanWKStefanWK

727




727












  • $begingroup$
    For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:47










  • $begingroup$
    It is the way around, see math.stackexchange.com/questions/1990936/….
    $endgroup$
    – Martin R
    Nov 27 '18 at 20:00










  • $begingroup$
    alright thanks for your help
    $endgroup$
    – StefanWK
    Nov 27 '18 at 20:16


















  • $begingroup$
    For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:47










  • $begingroup$
    It is the way around, see math.stackexchange.com/questions/1990936/….
    $endgroup$
    – Martin R
    Nov 27 '18 at 20:00










  • $begingroup$
    alright thanks for your help
    $endgroup$
    – StefanWK
    Nov 27 '18 at 20:16
















$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47




$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47












$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00




$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00












$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16




$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Either my above counter-example or your proof must be wrong...
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:48








  • 1




    $begingroup$
    Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
    $endgroup$
    – user376343
    Nov 27 '18 at 19:51










  • $begingroup$
    True enough, I'll change it!
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 21:47











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Either my above counter-example or your proof must be wrong...
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:48








  • 1




    $begingroup$
    Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
    $endgroup$
    – user376343
    Nov 27 '18 at 19:51










  • $begingroup$
    True enough, I'll change it!
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 21:47
















1












$begingroup$

The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Either my above counter-example or your proof must be wrong...
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:48








  • 1




    $begingroup$
    Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
    $endgroup$
    – user376343
    Nov 27 '18 at 19:51










  • $begingroup$
    True enough, I'll change it!
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 21:47














1












1








1





$begingroup$

The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.






share|cite|improve this answer











$endgroup$



The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 '18 at 21:48

























answered Nov 27 '18 at 19:38









Olivier MoschettaOlivier Moschetta

2,8311411




2,8311411












  • $begingroup$
    Either my above counter-example or your proof must be wrong...
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:48








  • 1




    $begingroup$
    Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
    $endgroup$
    – user376343
    Nov 27 '18 at 19:51










  • $begingroup$
    True enough, I'll change it!
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 21:47


















  • $begingroup$
    Either my above counter-example or your proof must be wrong...
    $endgroup$
    – Martin R
    Nov 27 '18 at 19:48








  • 1




    $begingroup$
    Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
    $endgroup$
    – user376343
    Nov 27 '18 at 19:51










  • $begingroup$
    True enough, I'll change it!
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 21:47
















$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48






$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48






1




1




$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51




$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51












$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47




$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47


















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