How to show $ax^t+by^tle (ax+by)^t.$












0












$begingroup$


Let $0<a,b,t<1$ and $x,y>0$. Show that
$$ax^t+by^tle (ax+by)^t.$$



I want to use this for a step in another proof. I've convinced myself it is correct since I have yet to find a counter example. My question is, how can I prove this inequality? I have tried directly and by contradiction, but I get no where.






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    @MostafaAyaz: Why did you change $ax^t+by^t$ to $(ax)^t + (by)^t$ ???
    $endgroup$
    – Martin R
    Jan 12 at 18:55










  • $begingroup$
    Because in that case the inequality doesn't hold.....
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 18:55






  • 1




    $begingroup$
    This is equivalent to $u^t+v^tle(u+v)^t$ with $tin(0,1)$ and $u,v>0$, isn't it? No sense in using five variables when three would do.
    $endgroup$
    – TonyK
    Jan 12 at 18:55












  • $begingroup$
    It seems to me you have the inequality flipped. The statement is true for $u^t + v^t geq (u + v)^t$, using @TonyK's notation.
    $endgroup$
    – parsiad
    Jan 12 at 19:00






  • 2




    $begingroup$
    I have rolled back the edit. It is not our task to guess what the real question is.
    $endgroup$
    – Martin R
    Jan 12 at 19:02
















0












$begingroup$


Let $0<a,b,t<1$ and $x,y>0$. Show that
$$ax^t+by^tle (ax+by)^t.$$



I want to use this for a step in another proof. I've convinced myself it is correct since I have yet to find a counter example. My question is, how can I prove this inequality? I have tried directly and by contradiction, but I get no where.






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    @MostafaAyaz: Why did you change $ax^t+by^t$ to $(ax)^t + (by)^t$ ???
    $endgroup$
    – Martin R
    Jan 12 at 18:55










  • $begingroup$
    Because in that case the inequality doesn't hold.....
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 18:55






  • 1




    $begingroup$
    This is equivalent to $u^t+v^tle(u+v)^t$ with $tin(0,1)$ and $u,v>0$, isn't it? No sense in using five variables when three would do.
    $endgroup$
    – TonyK
    Jan 12 at 18:55












  • $begingroup$
    It seems to me you have the inequality flipped. The statement is true for $u^t + v^t geq (u + v)^t$, using @TonyK's notation.
    $endgroup$
    – parsiad
    Jan 12 at 19:00






  • 2




    $begingroup$
    I have rolled back the edit. It is not our task to guess what the real question is.
    $endgroup$
    – Martin R
    Jan 12 at 19:02














0












0








0





$begingroup$


Let $0<a,b,t<1$ and $x,y>0$. Show that
$$ax^t+by^tle (ax+by)^t.$$



I want to use this for a step in another proof. I've convinced myself it is correct since I have yet to find a counter example. My question is, how can I prove this inequality? I have tried directly and by contradiction, but I get no where.






real-analysis







share|cite|improve this question











$endgroup$




Let $0<a,b,t<1$ and $x,y>0$. Show that
$$ax^t+by^tle (ax+by)^t.$$



I want to use this for a step in another proof. I've convinced myself it is correct since I have yet to find a counter example. My question is, how can I prove this inequality? I have tried directly and by contradiction, but I get no where.






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 19:02









Martin R

27.9k33255




27.9k33255










asked Jan 12 at 18:46









user906357user906357

1042




1042












  • $begingroup$
    @MostafaAyaz: Why did you change $ax^t+by^t$ to $(ax)^t + (by)^t$ ???
    $endgroup$
    – Martin R
    Jan 12 at 18:55










  • $begingroup$
    Because in that case the inequality doesn't hold.....
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 18:55






  • 1




    $begingroup$
    This is equivalent to $u^t+v^tle(u+v)^t$ with $tin(0,1)$ and $u,v>0$, isn't it? No sense in using five variables when three would do.
    $endgroup$
    – TonyK
    Jan 12 at 18:55












  • $begingroup$
    It seems to me you have the inequality flipped. The statement is true for $u^t + v^t geq (u + v)^t$, using @TonyK's notation.
    $endgroup$
    – parsiad
    Jan 12 at 19:00






  • 2




    $begingroup$
    I have rolled back the edit. It is not our task to guess what the real question is.
    $endgroup$
    – Martin R
    Jan 12 at 19:02


















  • $begingroup$
    @MostafaAyaz: Why did you change $ax^t+by^t$ to $(ax)^t + (by)^t$ ???
    $endgroup$
    – Martin R
    Jan 12 at 18:55










  • $begingroup$
    Because in that case the inequality doesn't hold.....
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 18:55






  • 1




    $begingroup$
    This is equivalent to $u^t+v^tle(u+v)^t$ with $tin(0,1)$ and $u,v>0$, isn't it? No sense in using five variables when three would do.
    $endgroup$
    – TonyK
    Jan 12 at 18:55












  • $begingroup$
    It seems to me you have the inequality flipped. The statement is true for $u^t + v^t geq (u + v)^t$, using @TonyK's notation.
    $endgroup$
    – parsiad
    Jan 12 at 19:00






  • 2




    $begingroup$
    I have rolled back the edit. It is not our task to guess what the real question is.
    $endgroup$
    – Martin R
    Jan 12 at 19:02
















$begingroup$
@MostafaAyaz: Why did you change $ax^t+by^t$ to $(ax)^t + (by)^t$ ???
$endgroup$
– Martin R
Jan 12 at 18:55




$begingroup$
@MostafaAyaz: Why did you change $ax^t+by^t$ to $(ax)^t + (by)^t$ ???
$endgroup$
– Martin R
Jan 12 at 18:55












$begingroup$
Because in that case the inequality doesn't hold.....
$endgroup$
– Mostafa Ayaz
Jan 12 at 18:55




$begingroup$
Because in that case the inequality doesn't hold.....
$endgroup$
– Mostafa Ayaz
Jan 12 at 18:55




1




1




$begingroup$
This is equivalent to $u^t+v^tle(u+v)^t$ with $tin(0,1)$ and $u,v>0$, isn't it? No sense in using five variables when three would do.
$endgroup$
– TonyK
Jan 12 at 18:55






$begingroup$
This is equivalent to $u^t+v^tle(u+v)^t$ with $tin(0,1)$ and $u,v>0$, isn't it? No sense in using five variables when three would do.
$endgroup$
– TonyK
Jan 12 at 18:55














$begingroup$
It seems to me you have the inequality flipped. The statement is true for $u^t + v^t geq (u + v)^t$, using @TonyK's notation.
$endgroup$
– parsiad
Jan 12 at 19:00




$begingroup$
It seems to me you have the inequality flipped. The statement is true for $u^t + v^t geq (u + v)^t$, using @TonyK's notation.
$endgroup$
– parsiad
Jan 12 at 19:00




2




2




$begingroup$
I have rolled back the edit. It is not our task to guess what the real question is.
$endgroup$
– Martin R
Jan 12 at 19:02




$begingroup$
I have rolled back the edit. It is not our task to guess what the real question is.
$endgroup$
– Martin R
Jan 12 at 19:02










1 Answer
1






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oldest

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0












$begingroup$

This expands on the comments, particularly by TonyK:



Let $tin(0,1)$, and set $f(x)=x^t$. Note that $f'(x)=tx^{t-1}$ and $f''(x)=t(t-1)x^{t-2}$. For $x>0$, note that $f''(x)<0$. Hence $f(x)$ is concave. Since also $f(0)ge 0$, this means that $f(x)$ is subadditive. This means that $$f(u+v)le f(u)+f(v)$$
Now, setting $u=ax, v=by$, we get
$$f(ax+by)le f(ax)+f(by)$$
or $$(ax+by)^tle (ax)^t+(by)^t$$



Comments: This only requires $a,x,b,y>0$, not necessarily in $(0,1)$. Also, this is reversed from the original inequality posted, and includes parentheses not present in the original question.






share|cite|improve this answer









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    0












    $begingroup$

    This expands on the comments, particularly by TonyK:



    Let $tin(0,1)$, and set $f(x)=x^t$. Note that $f'(x)=tx^{t-1}$ and $f''(x)=t(t-1)x^{t-2}$. For $x>0$, note that $f''(x)<0$. Hence $f(x)$ is concave. Since also $f(0)ge 0$, this means that $f(x)$ is subadditive. This means that $$f(u+v)le f(u)+f(v)$$
    Now, setting $u=ax, v=by$, we get
    $$f(ax+by)le f(ax)+f(by)$$
    or $$(ax+by)^tle (ax)^t+(by)^t$$



    Comments: This only requires $a,x,b,y>0$, not necessarily in $(0,1)$. Also, this is reversed from the original inequality posted, and includes parentheses not present in the original question.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This expands on the comments, particularly by TonyK:



      Let $tin(0,1)$, and set $f(x)=x^t$. Note that $f'(x)=tx^{t-1}$ and $f''(x)=t(t-1)x^{t-2}$. For $x>0$, note that $f''(x)<0$. Hence $f(x)$ is concave. Since also $f(0)ge 0$, this means that $f(x)$ is subadditive. This means that $$f(u+v)le f(u)+f(v)$$
      Now, setting $u=ax, v=by$, we get
      $$f(ax+by)le f(ax)+f(by)$$
      or $$(ax+by)^tle (ax)^t+(by)^t$$



      Comments: This only requires $a,x,b,y>0$, not necessarily in $(0,1)$. Also, this is reversed from the original inequality posted, and includes parentheses not present in the original question.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This expands on the comments, particularly by TonyK:



        Let $tin(0,1)$, and set $f(x)=x^t$. Note that $f'(x)=tx^{t-1}$ and $f''(x)=t(t-1)x^{t-2}$. For $x>0$, note that $f''(x)<0$. Hence $f(x)$ is concave. Since also $f(0)ge 0$, this means that $f(x)$ is subadditive. This means that $$f(u+v)le f(u)+f(v)$$
        Now, setting $u=ax, v=by$, we get
        $$f(ax+by)le f(ax)+f(by)$$
        or $$(ax+by)^tle (ax)^t+(by)^t$$



        Comments: This only requires $a,x,b,y>0$, not necessarily in $(0,1)$. Also, this is reversed from the original inequality posted, and includes parentheses not present in the original question.






        share|cite|improve this answer









        $endgroup$



        This expands on the comments, particularly by TonyK:



        Let $tin(0,1)$, and set $f(x)=x^t$. Note that $f'(x)=tx^{t-1}$ and $f''(x)=t(t-1)x^{t-2}$. For $x>0$, note that $f''(x)<0$. Hence $f(x)$ is concave. Since also $f(0)ge 0$, this means that $f(x)$ is subadditive. This means that $$f(u+v)le f(u)+f(v)$$
        Now, setting $u=ax, v=by$, we get
        $$f(ax+by)le f(ax)+f(by)$$
        or $$(ax+by)^tle (ax)^t+(by)^t$$



        Comments: This only requires $a,x,b,y>0$, not necessarily in $(0,1)$. Also, this is reversed from the original inequality posted, and includes parentheses not present in the original question.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 19:08









        vadim123vadim123

        75.9k897189




        75.9k897189






























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