Proof for $a^c + b^c > (a + b)^c$ when $0 < c 0$ and $1/c$ is nonintegral












2












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What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).



But what about the case when $1/c$ is non-integral?






algebra-precalculus inequality







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  • $begingroup$
    Start with $$dfrac ab=d$$
    $endgroup$
    – lab bhattacharjee
    Dec 30 '18 at 4:51






  • 2




    $begingroup$
    Is Jensen's inequality available to you?
    $endgroup$
    – DanielV
    Dec 30 '18 at 4:54






  • 3




    $begingroup$
    This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
    $endgroup$
    – Ben W
    Dec 30 '18 at 4:56
















2












$begingroup$


What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).



But what about the case when $1/c$ is non-integral?






algebra-precalculus inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Start with $$dfrac ab=d$$
    $endgroup$
    – lab bhattacharjee
    Dec 30 '18 at 4:51






  • 2




    $begingroup$
    Is Jensen's inequality available to you?
    $endgroup$
    – DanielV
    Dec 30 '18 at 4:54






  • 3




    $begingroup$
    This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
    $endgroup$
    – Ben W
    Dec 30 '18 at 4:56














2












2








2


2



$begingroup$


What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).



But what about the case when $1/c$ is non-integral?






algebra-precalculus inequality







share|cite|improve this question











$endgroup$




What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).



But what about the case when $1/c$ is non-integral?






algebra-precalculus inequality




algebra-precalculus inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 20:20









Eric Wofsey

181k12208336




181k12208336










asked Dec 30 '18 at 4:47









user1246462user1246462

1503




1503












  • $begingroup$
    Start with $$dfrac ab=d$$
    $endgroup$
    – lab bhattacharjee
    Dec 30 '18 at 4:51






  • 2




    $begingroup$
    Is Jensen's inequality available to you?
    $endgroup$
    – DanielV
    Dec 30 '18 at 4:54






  • 3




    $begingroup$
    This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
    $endgroup$
    – Ben W
    Dec 30 '18 at 4:56


















  • $begingroup$
    Start with $$dfrac ab=d$$
    $endgroup$
    – lab bhattacharjee
    Dec 30 '18 at 4:51






  • 2




    $begingroup$
    Is Jensen's inequality available to you?
    $endgroup$
    – DanielV
    Dec 30 '18 at 4:54






  • 3




    $begingroup$
    This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
    $endgroup$
    – Ben W
    Dec 30 '18 at 4:56
















$begingroup$
Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51




$begingroup$
Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51




2




2




$begingroup$
Is Jensen's inequality available to you?
$endgroup$
– DanielV
Dec 30 '18 at 4:54




$begingroup$
Is Jensen's inequality available to you?
$endgroup$
– DanielV
Dec 30 '18 at 4:54




3




3




$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56




$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56










2 Answers
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$(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$






    share|cite|improve this answer









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      2 Answers
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      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      3












      $begingroup$

      $(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        $(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          $(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.






          share|cite|improve this answer









          $endgroup$



          $(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 5:21









          Kavi Rama MurthyKavi Rama Murthy

          52.8k32055




          52.8k32055























              0












              $begingroup$

              The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$






                  share|cite|improve this answer









                  $endgroup$



                  The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 1 at 19:55









                  user376343user376343

                  3,3032825




                  3,3032825






























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