Prove inequality $y^(n-1)+x^nleq x^(n-1)+y^n$

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Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?










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  • Anything wrong with the question (because of the downvote)?
    – EuklidAlexandria
    Sep 10 at 16:06














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down vote

favorite












Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?










share|cite|improve this question





















  • Anything wrong with the question (because of the downvote)?
    – EuklidAlexandria
    Sep 10 at 16:06












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?










share|cite|improve this question













Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?







algebra-precalculus inequality






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asked Sep 10 at 8:35









EuklidAlexandria

3019




3019











  • Anything wrong with the question (because of the downvote)?
    – EuklidAlexandria
    Sep 10 at 16:06
















  • Anything wrong with the question (because of the downvote)?
    – EuklidAlexandria
    Sep 10 at 16:06















Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06




Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06










4 Answers
4






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oldest

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up vote
3
down vote



accepted










$f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.






share|cite|improve this answer



























    up vote
    3
    down vote













    $$1<x<yimplies 0<x-1<y-1.$$



    Multiply the left inequalities $n-1$ times and the right ones once to get



    $$x^n-1(x-1)<y^n-1(y-1).$$



    Equality is not possible.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Need to show:



      $x^n -x^n-1 lt y^n-y^n-1$;



      $x^n-1(x-1) lt y^n-1(y-1)$.



      Since $1<x<y$ :



      $(y-1)>(x-1)>0$, and



      $y^n-1 > x^n-1>0$(why?).






      share|cite|improve this answer



























        up vote
        2
        down vote













        The inequality is equivalent to
        $$
        x^n-1(x-1)leq y^n-1(y-1)
        $$
        This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).






        share|cite|improve this answer






















        • The method is good, but you are on the wrong side, $1<x<y$.
          – Yves Daoust
          Sep 10 at 8:50










        • @YvesDaoust It should be correct by now.
          – Janik
          Sep 10 at 8:55










        Your Answer




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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        $f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.






        share|cite|improve this answer
























          up vote
          3
          down vote



          accepted










          $f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.






          share|cite|improve this answer






















            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            $f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.






            share|cite|improve this answer












            $f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 10 at 8:40









            Kavi Rama Murthy

            27.1k31439




            27.1k31439




















                up vote
                3
                down vote













                $$1<x<yimplies 0<x-1<y-1.$$



                Multiply the left inequalities $n-1$ times and the right ones once to get



                $$x^n-1(x-1)<y^n-1(y-1).$$



                Equality is not possible.






                share|cite|improve this answer
























                  up vote
                  3
                  down vote













                  $$1<x<yimplies 0<x-1<y-1.$$



                  Multiply the left inequalities $n-1$ times and the right ones once to get



                  $$x^n-1(x-1)<y^n-1(y-1).$$



                  Equality is not possible.






                  share|cite|improve this answer






















                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    $$1<x<yimplies 0<x-1<y-1.$$



                    Multiply the left inequalities $n-1$ times and the right ones once to get



                    $$x^n-1(x-1)<y^n-1(y-1).$$



                    Equality is not possible.






                    share|cite|improve this answer












                    $$1<x<yimplies 0<x-1<y-1.$$



                    Multiply the left inequalities $n-1$ times and the right ones once to get



                    $$x^n-1(x-1)<y^n-1(y-1).$$



                    Equality is not possible.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 10 at 8:58









                    Yves Daoust

                    115k667210




                    115k667210




















                        up vote
                        2
                        down vote













                        Need to show:



                        $x^n -x^n-1 lt y^n-y^n-1$;



                        $x^n-1(x-1) lt y^n-1(y-1)$.



                        Since $1<x<y$ :



                        $(y-1)>(x-1)>0$, and



                        $y^n-1 > x^n-1>0$(why?).






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          Need to show:



                          $x^n -x^n-1 lt y^n-y^n-1$;



                          $x^n-1(x-1) lt y^n-1(y-1)$.



                          Since $1<x<y$ :



                          $(y-1)>(x-1)>0$, and



                          $y^n-1 > x^n-1>0$(why?).






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            Need to show:



                            $x^n -x^n-1 lt y^n-y^n-1$;



                            $x^n-1(x-1) lt y^n-1(y-1)$.



                            Since $1<x<y$ :



                            $(y-1)>(x-1)>0$, and



                            $y^n-1 > x^n-1>0$(why?).






                            share|cite|improve this answer












                            Need to show:



                            $x^n -x^n-1 lt y^n-y^n-1$;



                            $x^n-1(x-1) lt y^n-1(y-1)$.



                            Since $1<x<y$ :



                            $(y-1)>(x-1)>0$, and



                            $y^n-1 > x^n-1>0$(why?).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 10 at 8:52









                            Peter Szilas

                            8,4452617




                            8,4452617




















                                up vote
                                2
                                down vote













                                The inequality is equivalent to
                                $$
                                x^n-1(x-1)leq y^n-1(y-1)
                                $$
                                This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).






                                share|cite|improve this answer






















                                • The method is good, but you are on the wrong side, $1<x<y$.
                                  – Yves Daoust
                                  Sep 10 at 8:50










                                • @YvesDaoust It should be correct by now.
                                  – Janik
                                  Sep 10 at 8:55














                                up vote
                                2
                                down vote













                                The inequality is equivalent to
                                $$
                                x^n-1(x-1)leq y^n-1(y-1)
                                $$
                                This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).






                                share|cite|improve this answer






















                                • The method is good, but you are on the wrong side, $1<x<y$.
                                  – Yves Daoust
                                  Sep 10 at 8:50










                                • @YvesDaoust It should be correct by now.
                                  – Janik
                                  Sep 10 at 8:55












                                up vote
                                2
                                down vote










                                up vote
                                2
                                down vote









                                The inequality is equivalent to
                                $$
                                x^n-1(x-1)leq y^n-1(y-1)
                                $$
                                This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).






                                share|cite|improve this answer














                                The inequality is equivalent to
                                $$
                                x^n-1(x-1)leq y^n-1(y-1)
                                $$
                                This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Sep 10 at 8:54

























                                answered Sep 10 at 8:43









                                Janik

                                1,4352418




                                1,4352418











                                • The method is good, but you are on the wrong side, $1<x<y$.
                                  – Yves Daoust
                                  Sep 10 at 8:50










                                • @YvesDaoust It should be correct by now.
                                  – Janik
                                  Sep 10 at 8:55
















                                • The method is good, but you are on the wrong side, $1<x<y$.
                                  – Yves Daoust
                                  Sep 10 at 8:50










                                • @YvesDaoust It should be correct by now.
                                  – Janik
                                  Sep 10 at 8:55















                                The method is good, but you are on the wrong side, $1<x<y$.
                                – Yves Daoust
                                Sep 10 at 8:50




                                The method is good, but you are on the wrong side, $1<x<y$.
                                – Yves Daoust
                                Sep 10 at 8:50












                                @YvesDaoust It should be correct by now.
                                – Janik
                                Sep 10 at 8:55




                                @YvesDaoust It should be correct by now.
                                – Janik
                                Sep 10 at 8:55

















                                 

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