Prove the inequality $(1 + x)^{1/pi} 0$












1












$begingroup$



Prove the inequality $(1 + x)^{1/pi} < 1 + frac{x}{pi}$ for $x > 0$




The inequality looks very straightforward to prove; however, I have been struggling for a while. The first thing that came to my mind was the Binomial Series, but there are both positive and negative terms in there, so it is not clear that the result follows.



I also tried defining a function $f(x)$ and taking derivatives, but that's not valid since the power rule only holds for rationals and $1/pi$ is irrational.



EDIT: this also looks a lot like Bernoulli's inequality; however, I think the exponent has to be an integer in that.




Let $f(x) = frac{1}{pi}log(1 + x) - log(1 + frac{x}{pi})$, and note that $f(0) = 0$. So, by showing that $f$ is decreasing, our result will follow. We have



$$f'(x) = frac{1}{pi} frac{1}{1 + x} -frac{1}{1 + frac{x}{pi}} cdot frac{1}{pi} $$



$$ frac{1}{pi}left(frac{1}{1 + x} - frac{1}{1 + frac{x}{pi}}right) < 0 $$



$$Longleftrightarrow frac{1}{1 + x} < frac{1}{1 + frac{x}{pi}} $$



$$ Longleftrightarrow 1 + x > 1 + x/pi $$



$$x > x/pi, $$



which is clearly true since $x > 0$.



So, since we have shown that the logarithm of the function is negative for $x > 0$, it follows that the function itself is decreasing.






real-analysis inequality real-numbers







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  • 1




    $begingroup$
    If I'm not wrong, it's "Bernoulli" and not "Beronulli"
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 15:43






  • 1




    $begingroup$
    Power rule is valid for all real powers
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 15:43










  • $begingroup$
    What precise statement do you call "Bernoulli's Inequality"? There's a very simple one and a very sweeping one.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 15:48










  • $begingroup$
    is my original post attempt correct ?
    $endgroup$
    – joseph
    Dec 14 '18 at 15:56










  • $begingroup$
    You should indicate the editing when the nature of the question change completely, even for a typo.
    $endgroup$
    – gimusi
    Dec 14 '18 at 16:02
















1












$begingroup$



Prove the inequality $(1 + x)^{1/pi} < 1 + frac{x}{pi}$ for $x > 0$




The inequality looks very straightforward to prove; however, I have been struggling for a while. The first thing that came to my mind was the Binomial Series, but there are both positive and negative terms in there, so it is not clear that the result follows.



I also tried defining a function $f(x)$ and taking derivatives, but that's not valid since the power rule only holds for rationals and $1/pi$ is irrational.



EDIT: this also looks a lot like Bernoulli's inequality; however, I think the exponent has to be an integer in that.




Let $f(x) = frac{1}{pi}log(1 + x) - log(1 + frac{x}{pi})$, and note that $f(0) = 0$. So, by showing that $f$ is decreasing, our result will follow. We have



$$f'(x) = frac{1}{pi} frac{1}{1 + x} -frac{1}{1 + frac{x}{pi}} cdot frac{1}{pi} $$



$$ frac{1}{pi}left(frac{1}{1 + x} - frac{1}{1 + frac{x}{pi}}right) < 0 $$



$$Longleftrightarrow frac{1}{1 + x} < frac{1}{1 + frac{x}{pi}} $$



$$ Longleftrightarrow 1 + x > 1 + x/pi $$



$$x > x/pi, $$



which is clearly true since $x > 0$.



So, since we have shown that the logarithm of the function is negative for $x > 0$, it follows that the function itself is decreasing.






real-analysis inequality real-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If I'm not wrong, it's "Bernoulli" and not "Beronulli"
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 15:43






  • 1




    $begingroup$
    Power rule is valid for all real powers
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 15:43










  • $begingroup$
    What precise statement do you call "Bernoulli's Inequality"? There's a very simple one and a very sweeping one.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 15:48










  • $begingroup$
    is my original post attempt correct ?
    $endgroup$
    – joseph
    Dec 14 '18 at 15:56










  • $begingroup$
    You should indicate the editing when the nature of the question change completely, even for a typo.
    $endgroup$
    – gimusi
    Dec 14 '18 at 16:02














1












1








1


1



$begingroup$



Prove the inequality $(1 + x)^{1/pi} < 1 + frac{x}{pi}$ for $x > 0$




The inequality looks very straightforward to prove; however, I have been struggling for a while. The first thing that came to my mind was the Binomial Series, but there are both positive and negative terms in there, so it is not clear that the result follows.



I also tried defining a function $f(x)$ and taking derivatives, but that's not valid since the power rule only holds for rationals and $1/pi$ is irrational.



EDIT: this also looks a lot like Bernoulli's inequality; however, I think the exponent has to be an integer in that.




Let $f(x) = frac{1}{pi}log(1 + x) - log(1 + frac{x}{pi})$, and note that $f(0) = 0$. So, by showing that $f$ is decreasing, our result will follow. We have



$$f'(x) = frac{1}{pi} frac{1}{1 + x} -frac{1}{1 + frac{x}{pi}} cdot frac{1}{pi} $$



$$ frac{1}{pi}left(frac{1}{1 + x} - frac{1}{1 + frac{x}{pi}}right) < 0 $$



$$Longleftrightarrow frac{1}{1 + x} < frac{1}{1 + frac{x}{pi}} $$



$$ Longleftrightarrow 1 + x > 1 + x/pi $$



$$x > x/pi, $$



which is clearly true since $x > 0$.



So, since we have shown that the logarithm of the function is negative for $x > 0$, it follows that the function itself is decreasing.






real-analysis inequality real-numbers







share|cite|improve this question











$endgroup$





Prove the inequality $(1 + x)^{1/pi} < 1 + frac{x}{pi}$ for $x > 0$




The inequality looks very straightforward to prove; however, I have been struggling for a while. The first thing that came to my mind was the Binomial Series, but there are both positive and negative terms in there, so it is not clear that the result follows.



I also tried defining a function $f(x)$ and taking derivatives, but that's not valid since the power rule only holds for rationals and $1/pi$ is irrational.



EDIT: this also looks a lot like Bernoulli's inequality; however, I think the exponent has to be an integer in that.




Let $f(x) = frac{1}{pi}log(1 + x) - log(1 + frac{x}{pi})$, and note that $f(0) = 0$. So, by showing that $f$ is decreasing, our result will follow. We have



$$f'(x) = frac{1}{pi} frac{1}{1 + x} -frac{1}{1 + frac{x}{pi}} cdot frac{1}{pi} $$



$$ frac{1}{pi}left(frac{1}{1 + x} - frac{1}{1 + frac{x}{pi}}right) < 0 $$



$$Longleftrightarrow frac{1}{1 + x} < frac{1}{1 + frac{x}{pi}} $$



$$ Longleftrightarrow 1 + x > 1 + x/pi $$



$$x > x/pi, $$



which is clearly true since $x > 0$.



So, since we have shown that the logarithm of the function is negative for $x > 0$, it follows that the function itself is decreasing.






real-analysis inequality real-numbers




real-analysis inequality real-numbers






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 16:06







joseph

















asked Dec 14 '18 at 15:38









josephjoseph

496111




496111








  • 1




    $begingroup$
    If I'm not wrong, it's "Bernoulli" and not "Beronulli"
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 15:43






  • 1




    $begingroup$
    Power rule is valid for all real powers
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 15:43










  • $begingroup$
    What precise statement do you call "Bernoulli's Inequality"? There's a very simple one and a very sweeping one.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 15:48










  • $begingroup$
    is my original post attempt correct ?
    $endgroup$
    – joseph
    Dec 14 '18 at 15:56










  • $begingroup$
    You should indicate the editing when the nature of the question change completely, even for a typo.
    $endgroup$
    – gimusi
    Dec 14 '18 at 16:02














  • 1




    $begingroup$
    If I'm not wrong, it's "Bernoulli" and not "Beronulli"
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 15:43






  • 1




    $begingroup$
    Power rule is valid for all real powers
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 15:43










  • $begingroup$
    What precise statement do you call "Bernoulli's Inequality"? There's a very simple one and a very sweeping one.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 15:48










  • $begingroup$
    is my original post attempt correct ?
    $endgroup$
    – joseph
    Dec 14 '18 at 15:56










  • $begingroup$
    You should indicate the editing when the nature of the question change completely, even for a typo.
    $endgroup$
    – gimusi
    Dec 14 '18 at 16:02








1




1




$begingroup$
If I'm not wrong, it's "Bernoulli" and not "Beronulli"
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:43




$begingroup$
If I'm not wrong, it's "Bernoulli" and not "Beronulli"
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:43




1




1




$begingroup$
Power rule is valid for all real powers
$endgroup$
– Shubham Johri
Dec 14 '18 at 15:43




$begingroup$
Power rule is valid for all real powers
$endgroup$
– Shubham Johri
Dec 14 '18 at 15:43












$begingroup$
What precise statement do you call "Bernoulli's Inequality"? There's a very simple one and a very sweeping one.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 15:48




$begingroup$
What precise statement do you call "Bernoulli's Inequality"? There's a very simple one and a very sweeping one.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 15:48












$begingroup$
is my original post attempt correct ?
$endgroup$
– joseph
Dec 14 '18 at 15:56




$begingroup$
is my original post attempt correct ?
$endgroup$
– joseph
Dec 14 '18 at 15:56












$begingroup$
You should indicate the editing when the nature of the question change completely, even for a typo.
$endgroup$
– gimusi
Dec 14 '18 at 16:02




$begingroup$
You should indicate the editing when the nature of the question change completely, even for a typo.
$endgroup$
– gimusi
Dec 14 '18 at 16:02










4 Answers
4






active

oldest

votes


















2












$begingroup$

Answer to the original question: $(1 + x)^{1/pi} > 1 + frac{x}{pi}, , x>0$



The inequality is not true, indeed for $x=1$ we have



$$(1 + 1)^{1/pi}approx 1.247$$



$$1 + frac{1}{pi}approx 1.318$$



indeed for $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



$$(1+x)^r le 1+rx$$



Answer after editing: $(1 + x)^{1/pi} < 1 + frac{x}{pi}, , x>0$



For $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



$$(1+x)^r le 1+rx$$



and equality holds only for $r=0,1$ or $x=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    does my way work too ?
    $endgroup$
    – joseph
    Dec 14 '18 at 16:04










  • $begingroup$
    @joseph I'll check that.
    $endgroup$
    – gimusi
    Dec 14 '18 at 16:05










  • $begingroup$
    @joseph Yes it looks fine!
    $endgroup$
    – gimusi
    Dec 14 '18 at 16:18



















4












$begingroup$

Defining $$f(x)=frac{1}{pi}ln(1+x)-lnleft(1+frac{x}{pi}right)$$ and now use calculus. For $$x>0$$ we get $$xleft(1-frac{1}{pi}right)>0$$ which is true.
$$x=0$$ gives the maximum of our function. The first derivative of our $$f(x)$$ is given by
$$frac{1}{pi(1+x)}-frac{1}{1+frac{x}{pi}}cdot frac{1}{pi}$$
solving $$f'(x)=0$$ we get $$1+frac{x}{pi}-1-x=0$$ and $$frac{x}{pi}-x=0$$ is $$xleft(frac{1}{pi}-1right)=0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    hi, i tried your way. i edited my original post. can you check it please
    $endgroup$
    – joseph
    Dec 14 '18 at 15:52










  • $begingroup$
    how did you get $x(1 - 1/pi)$?
    $endgroup$
    – joseph
    Dec 14 '18 at 16:08



















1












$begingroup$

The inequality is not true, because for $0<r<1$ we have that
$$(1+x)^r<1+rx$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Power rule is valid for all real powers.



    Define $f(x)=(1+x)^p-1-px; 0<p<1, xge0$



    $displaystyle f(0)=0, f'(x)=p(1+x)^{p-1}-p=frac p{(1+x)^{1-p}}-ple0$



    Therefore, $f(x)le0, forall xge0$






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Answer to the original question: $(1 + x)^{1/pi} > 1 + frac{x}{pi}, , x>0$



      The inequality is not true, indeed for $x=1$ we have



      $$(1 + 1)^{1/pi}approx 1.247$$



      $$1 + frac{1}{pi}approx 1.318$$



      indeed for $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      Answer after editing: $(1 + x)^{1/pi} < 1 + frac{x}{pi}, , x>0$



      For $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      and equality holds only for $r=0,1$ or $x=0$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        does my way work too ?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:04










      • $begingroup$
        @joseph I'll check that.
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:05










      • $begingroup$
        @joseph Yes it looks fine!
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:18
















      2












      $begingroup$

      Answer to the original question: $(1 + x)^{1/pi} > 1 + frac{x}{pi}, , x>0$



      The inequality is not true, indeed for $x=1$ we have



      $$(1 + 1)^{1/pi}approx 1.247$$



      $$1 + frac{1}{pi}approx 1.318$$



      indeed for $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      Answer after editing: $(1 + x)^{1/pi} < 1 + frac{x}{pi}, , x>0$



      For $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      and equality holds only for $r=0,1$ or $x=0$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        does my way work too ?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:04










      • $begingroup$
        @joseph I'll check that.
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:05










      • $begingroup$
        @joseph Yes it looks fine!
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:18














      2












      2








      2





      $begingroup$

      Answer to the original question: $(1 + x)^{1/pi} > 1 + frac{x}{pi}, , x>0$



      The inequality is not true, indeed for $x=1$ we have



      $$(1 + 1)^{1/pi}approx 1.247$$



      $$1 + frac{1}{pi}approx 1.318$$



      indeed for $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      Answer after editing: $(1 + x)^{1/pi} < 1 + frac{x}{pi}, , x>0$



      For $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      and equality holds only for $r=0,1$ or $x=0$.






      share|cite|improve this answer











      $endgroup$



      Answer to the original question: $(1 + x)^{1/pi} > 1 + frac{x}{pi}, , x>0$



      The inequality is not true, indeed for $x=1$ we have



      $$(1 + 1)^{1/pi}approx 1.247$$



      $$1 + frac{1}{pi}approx 1.318$$



      indeed for $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      Answer after editing: $(1 + x)^{1/pi} < 1 + frac{x}{pi}, , x>0$



      For $0le rle 1$ the following Bernoulli's inequality holds as $xge -1$



      $$(1+x)^r le 1+rx$$



      and equality holds only for $r=0,1$ or $x=0$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 14 '18 at 16:04

























      answered Dec 14 '18 at 15:42









      gimusigimusi

      92.9k84494




      92.9k84494












      • $begingroup$
        does my way work too ?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:04










      • $begingroup$
        @joseph I'll check that.
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:05










      • $begingroup$
        @joseph Yes it looks fine!
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:18


















      • $begingroup$
        does my way work too ?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:04










      • $begingroup$
        @joseph I'll check that.
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:05










      • $begingroup$
        @joseph Yes it looks fine!
        $endgroup$
        – gimusi
        Dec 14 '18 at 16:18
















      $begingroup$
      does my way work too ?
      $endgroup$
      – joseph
      Dec 14 '18 at 16:04




      $begingroup$
      does my way work too ?
      $endgroup$
      – joseph
      Dec 14 '18 at 16:04












      $begingroup$
      @joseph I'll check that.
      $endgroup$
      – gimusi
      Dec 14 '18 at 16:05




      $begingroup$
      @joseph I'll check that.
      $endgroup$
      – gimusi
      Dec 14 '18 at 16:05












      $begingroup$
      @joseph Yes it looks fine!
      $endgroup$
      – gimusi
      Dec 14 '18 at 16:18




      $begingroup$
      @joseph Yes it looks fine!
      $endgroup$
      – gimusi
      Dec 14 '18 at 16:18











      4












      $begingroup$

      Defining $$f(x)=frac{1}{pi}ln(1+x)-lnleft(1+frac{x}{pi}right)$$ and now use calculus. For $$x>0$$ we get $$xleft(1-frac{1}{pi}right)>0$$ which is true.
      $$x=0$$ gives the maximum of our function. The first derivative of our $$f(x)$$ is given by
      $$frac{1}{pi(1+x)}-frac{1}{1+frac{x}{pi}}cdot frac{1}{pi}$$
      solving $$f'(x)=0$$ we get $$1+frac{x}{pi}-1-x=0$$ and $$frac{x}{pi}-x=0$$ is $$xleft(frac{1}{pi}-1right)=0$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        hi, i tried your way. i edited my original post. can you check it please
        $endgroup$
        – joseph
        Dec 14 '18 at 15:52










      • $begingroup$
        how did you get $x(1 - 1/pi)$?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:08
















      4












      $begingroup$

      Defining $$f(x)=frac{1}{pi}ln(1+x)-lnleft(1+frac{x}{pi}right)$$ and now use calculus. For $$x>0$$ we get $$xleft(1-frac{1}{pi}right)>0$$ which is true.
      $$x=0$$ gives the maximum of our function. The first derivative of our $$f(x)$$ is given by
      $$frac{1}{pi(1+x)}-frac{1}{1+frac{x}{pi}}cdot frac{1}{pi}$$
      solving $$f'(x)=0$$ we get $$1+frac{x}{pi}-1-x=0$$ and $$frac{x}{pi}-x=0$$ is $$xleft(frac{1}{pi}-1right)=0$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        hi, i tried your way. i edited my original post. can you check it please
        $endgroup$
        – joseph
        Dec 14 '18 at 15:52










      • $begingroup$
        how did you get $x(1 - 1/pi)$?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:08














      4












      4








      4





      $begingroup$

      Defining $$f(x)=frac{1}{pi}ln(1+x)-lnleft(1+frac{x}{pi}right)$$ and now use calculus. For $$x>0$$ we get $$xleft(1-frac{1}{pi}right)>0$$ which is true.
      $$x=0$$ gives the maximum of our function. The first derivative of our $$f(x)$$ is given by
      $$frac{1}{pi(1+x)}-frac{1}{1+frac{x}{pi}}cdot frac{1}{pi}$$
      solving $$f'(x)=0$$ we get $$1+frac{x}{pi}-1-x=0$$ and $$frac{x}{pi}-x=0$$ is $$xleft(frac{1}{pi}-1right)=0$$






      share|cite|improve this answer











      $endgroup$



      Defining $$f(x)=frac{1}{pi}ln(1+x)-lnleft(1+frac{x}{pi}right)$$ and now use calculus. For $$x>0$$ we get $$xleft(1-frac{1}{pi}right)>0$$ which is true.
      $$x=0$$ gives the maximum of our function. The first derivative of our $$f(x)$$ is given by
      $$frac{1}{pi(1+x)}-frac{1}{1+frac{x}{pi}}cdot frac{1}{pi}$$
      solving $$f'(x)=0$$ we get $$1+frac{x}{pi}-1-x=0$$ and $$frac{x}{pi}-x=0$$ is $$xleft(frac{1}{pi}-1right)=0$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 14 '18 at 16:14

























      answered Dec 14 '18 at 15:42









      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

      76.8k42866




      76.8k42866












      • $begingroup$
        hi, i tried your way. i edited my original post. can you check it please
        $endgroup$
        – joseph
        Dec 14 '18 at 15:52










      • $begingroup$
        how did you get $x(1 - 1/pi)$?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:08


















      • $begingroup$
        hi, i tried your way. i edited my original post. can you check it please
        $endgroup$
        – joseph
        Dec 14 '18 at 15:52










      • $begingroup$
        how did you get $x(1 - 1/pi)$?
        $endgroup$
        – joseph
        Dec 14 '18 at 16:08
















      $begingroup$
      hi, i tried your way. i edited my original post. can you check it please
      $endgroup$
      – joseph
      Dec 14 '18 at 15:52




      $begingroup$
      hi, i tried your way. i edited my original post. can you check it please
      $endgroup$
      – joseph
      Dec 14 '18 at 15:52












      $begingroup$
      how did you get $x(1 - 1/pi)$?
      $endgroup$
      – joseph
      Dec 14 '18 at 16:08




      $begingroup$
      how did you get $x(1 - 1/pi)$?
      $endgroup$
      – joseph
      Dec 14 '18 at 16:08











      1












      $begingroup$

      The inequality is not true, because for $0<r<1$ we have that
      $$(1+x)^r<1+rx$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The inequality is not true, because for $0<r<1$ we have that
        $$(1+x)^r<1+rx$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The inequality is not true, because for $0<r<1$ we have that
          $$(1+x)^r<1+rx$$






          share|cite|improve this answer









          $endgroup$



          The inequality is not true, because for $0<r<1$ we have that
          $$(1+x)^r<1+rx$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 15:42









          BotondBotond

          5,9252832




          5,9252832























              0












              $begingroup$

              Power rule is valid for all real powers.



              Define $f(x)=(1+x)^p-1-px; 0<p<1, xge0$



              $displaystyle f(0)=0, f'(x)=p(1+x)^{p-1}-p=frac p{(1+x)^{1-p}}-ple0$



              Therefore, $f(x)le0, forall xge0$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Power rule is valid for all real powers.



                Define $f(x)=(1+x)^p-1-px; 0<p<1, xge0$



                $displaystyle f(0)=0, f'(x)=p(1+x)^{p-1}-p=frac p{(1+x)^{1-p}}-ple0$



                Therefore, $f(x)le0, forall xge0$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Power rule is valid for all real powers.



                  Define $f(x)=(1+x)^p-1-px; 0<p<1, xge0$



                  $displaystyle f(0)=0, f'(x)=p(1+x)^{p-1}-p=frac p{(1+x)^{1-p}}-ple0$



                  Therefore, $f(x)le0, forall xge0$






                  share|cite|improve this answer









                  $endgroup$



                  Power rule is valid for all real powers.



                  Define $f(x)=(1+x)^p-1-px; 0<p<1, xge0$



                  $displaystyle f(0)=0, f'(x)=p(1+x)^{p-1}-p=frac p{(1+x)^{1-p}}-ple0$



                  Therefore, $f(x)le0, forall xge0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 15:49









                  Shubham JohriShubham Johri

                  5,204718




                  5,204718






























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