Show that $a^{2014}+b^{2014}geq a^{2013}+b^{2013} $.












4












$begingroup$


Let $ a, bin mathbb {R}_{+} $ s.t. $a^{22}+b^{22}=a^{3}+b^{3} $.



Show that $a^{2014}+b^{2014}geq a^{2013}+b^{2013} $.



By Chebyshev's inequality we obtain $a^{19}+b^{19}leq 2Leftrightarrow b^{19}-1leq 1-a^{19}Leftrightarrow (b-1)(b^{18}+b^{17}+...+b+1)leq (1-a)(a^{18}+a^{17}+...+a+1) $.



I supposed $bgeq 1geq a $. Then $ b^{18}+b^{17}+...+b+1geq a^{18}+a^{17}+...+a+1$. Then $b-1leq 1-a Leftrightarrow a+bleq 2$. Now I am stuck.






inequality a.m.-g.m.-inequality holder-inequality tangent-line-method







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Let $ a, bin mathbb {R}_{+} $ s.t. $a^{22}+b^{22}=a^{3}+b^{3} $.



    Show that $a^{2014}+b^{2014}geq a^{2013}+b^{2013} $.



    By Chebyshev's inequality we obtain $a^{19}+b^{19}leq 2Leftrightarrow b^{19}-1leq 1-a^{19}Leftrightarrow (b-1)(b^{18}+b^{17}+...+b+1)leq (1-a)(a^{18}+a^{17}+...+a+1) $.



    I supposed $bgeq 1geq a $. Then $ b^{18}+b^{17}+...+b+1geq a^{18}+a^{17}+...+a+1$. Then $b-1leq 1-a Leftrightarrow a+bleq 2$. Now I am stuck.






    inequality a.m.-g.m.-inequality holder-inequality tangent-line-method







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      Let $ a, bin mathbb {R}_{+} $ s.t. $a^{22}+b^{22}=a^{3}+b^{3} $.



      Show that $a^{2014}+b^{2014}geq a^{2013}+b^{2013} $.



      By Chebyshev's inequality we obtain $a^{19}+b^{19}leq 2Leftrightarrow b^{19}-1leq 1-a^{19}Leftrightarrow (b-1)(b^{18}+b^{17}+...+b+1)leq (1-a)(a^{18}+a^{17}+...+a+1) $.



      I supposed $bgeq 1geq a $. Then $ b^{18}+b^{17}+...+b+1geq a^{18}+a^{17}+...+a+1$. Then $b-1leq 1-a Leftrightarrow a+bleq 2$. Now I am stuck.






      inequality a.m.-g.m.-inequality holder-inequality tangent-line-method







      share|cite|improve this question











      $endgroup$




      Let $ a, bin mathbb {R}_{+} $ s.t. $a^{22}+b^{22}=a^{3}+b^{3} $.



      Show that $a^{2014}+b^{2014}geq a^{2013}+b^{2013} $.



      By Chebyshev's inequality we obtain $a^{19}+b^{19}leq 2Leftrightarrow b^{19}-1leq 1-a^{19}Leftrightarrow (b-1)(b^{18}+b^{17}+...+b+1)leq (1-a)(a^{18}+a^{17}+...+a+1) $.



      I supposed $bgeq 1geq a $. Then $ b^{18}+b^{17}+...+b+1geq a^{18}+a^{17}+...+a+1$. Then $b-1leq 1-a Leftrightarrow a+bleq 2$. Now I am stuck.






      inequality a.m.-g.m.-inequality holder-inequality tangent-line-method




      inequality a.m.-g.m.-inequality holder-inequality tangent-line-method






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 21:25









      Michael Rozenberg

      108k1895200




      108k1895200










      asked Jan 26 at 19:26









      ProblemsolvingProblemsolving

      917412




      917412






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          Suppose $a, bne 1$. Let $f(x)=a^x+b^x$. The function is continuous, so its derivative has a root in $[3, 22]$. For $x_0$ to be a root of the derivative, the equation $$left(frac abright)^{x_0}=-frac{ln{b}}{ln{a}}$$ must hold. We can see that $x_0$ is unique, so the only root of $f'$ is in $[3, 22]$. Since for $xrightarrowinfty$ we have $f(x)rightarrowinfty$ (at least one of $a, b$ is $>1$), the derivative is positive for $x>22$, so $f(m)>f(n)$ for all $m>n>22$. In the case $m=2014$ and $n=2013$, we get the above inequality.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            There is an elementary proof?
            $endgroup$
            – Problemsolving
            Jan 26 at 20:57










          • $begingroup$
            There are some more elementary proofs (like the ones below), but this one proves a more general statement.
            $endgroup$
            – L3435
            Jan 27 at 20:43



















          1












          $begingroup$

          Let $f(x)=a^x+b^x , f(3)=f(22)$



          By Holder inequality:



          $$ left( f(3) right )^{frac{1991}{2010}}cdot (f(2013))^{1-frac{1991}{2010}} ge f(22) Rightarrow f(2013)ge f(22)=f(3)$$



          $$left( f(22 right )^{frac{1}{1992}}cdot (f(2014))^{1-frac{1}{1992}} ge f(2013) Rightarrow f(2014) ge f(2013)$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            We need to prove that
            $$sum_{cyc}left(a^{2014}-a^{2013}right)geq0$$ or
            $$sum_{cyc}left(a^{2014}-a^{2013}-frac{1}{19}left(a^{22}-a^3right)right)geq0$$ or
            $$sum_{cyc}a^3left(19a^{2011}-19a^{2010}-a^{19}+1right)geq0.$$
            Now, by AM-GM
            $$2010a^{2011}-2011a^{2010}+1geq0$$ or
            $$19a^{2011}-frac{19cdot2011}{2010}a^{2010}+frac{19}{2010}geq0.$$
            Id est, it's enough to prove that
            $$left(frac{19cdot2011}{2010}-19right)a^{2010}-a^{19}+1-frac{19}{2010}geq0$$ or
            $$19a^{2010}-2010a^{19}+1991geq0,$$ which is AM-GM again.



            Done!






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Suppose $a, bne 1$. Let $f(x)=a^x+b^x$. The function is continuous, so its derivative has a root in $[3, 22]$. For $x_0$ to be a root of the derivative, the equation $$left(frac abright)^{x_0}=-frac{ln{b}}{ln{a}}$$ must hold. We can see that $x_0$ is unique, so the only root of $f'$ is in $[3, 22]$. Since for $xrightarrowinfty$ we have $f(x)rightarrowinfty$ (at least one of $a, b$ is $>1$), the derivative is positive for $x>22$, so $f(m)>f(n)$ for all $m>n>22$. In the case $m=2014$ and $n=2013$, we get the above inequality.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                There is an elementary proof?
                $endgroup$
                – Problemsolving
                Jan 26 at 20:57










              • $begingroup$
                There are some more elementary proofs (like the ones below), but this one proves a more general statement.
                $endgroup$
                – L3435
                Jan 27 at 20:43
















              3












              $begingroup$

              Suppose $a, bne 1$. Let $f(x)=a^x+b^x$. The function is continuous, so its derivative has a root in $[3, 22]$. For $x_0$ to be a root of the derivative, the equation $$left(frac abright)^{x_0}=-frac{ln{b}}{ln{a}}$$ must hold. We can see that $x_0$ is unique, so the only root of $f'$ is in $[3, 22]$. Since for $xrightarrowinfty$ we have $f(x)rightarrowinfty$ (at least one of $a, b$ is $>1$), the derivative is positive for $x>22$, so $f(m)>f(n)$ for all $m>n>22$. In the case $m=2014$ and $n=2013$, we get the above inequality.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                There is an elementary proof?
                $endgroup$
                – Problemsolving
                Jan 26 at 20:57










              • $begingroup$
                There are some more elementary proofs (like the ones below), but this one proves a more general statement.
                $endgroup$
                – L3435
                Jan 27 at 20:43














              3












              3








              3





              $begingroup$

              Suppose $a, bne 1$. Let $f(x)=a^x+b^x$. The function is continuous, so its derivative has a root in $[3, 22]$. For $x_0$ to be a root of the derivative, the equation $$left(frac abright)^{x_0}=-frac{ln{b}}{ln{a}}$$ must hold. We can see that $x_0$ is unique, so the only root of $f'$ is in $[3, 22]$. Since for $xrightarrowinfty$ we have $f(x)rightarrowinfty$ (at least one of $a, b$ is $>1$), the derivative is positive for $x>22$, so $f(m)>f(n)$ for all $m>n>22$. In the case $m=2014$ and $n=2013$, we get the above inequality.






              share|cite|improve this answer









              $endgroup$



              Suppose $a, bne 1$. Let $f(x)=a^x+b^x$. The function is continuous, so its derivative has a root in $[3, 22]$. For $x_0$ to be a root of the derivative, the equation $$left(frac abright)^{x_0}=-frac{ln{b}}{ln{a}}$$ must hold. We can see that $x_0$ is unique, so the only root of $f'$ is in $[3, 22]$. Since for $xrightarrowinfty$ we have $f(x)rightarrowinfty$ (at least one of $a, b$ is $>1$), the derivative is positive for $x>22$, so $f(m)>f(n)$ for all $m>n>22$. In the case $m=2014$ and $n=2013$, we get the above inequality.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 26 at 20:32









              L3435L3435

              312




              312












              • $begingroup$
                There is an elementary proof?
                $endgroup$
                – Problemsolving
                Jan 26 at 20:57










              • $begingroup$
                There are some more elementary proofs (like the ones below), but this one proves a more general statement.
                $endgroup$
                – L3435
                Jan 27 at 20:43


















              • $begingroup$
                There is an elementary proof?
                $endgroup$
                – Problemsolving
                Jan 26 at 20:57










              • $begingroup$
                There are some more elementary proofs (like the ones below), but this one proves a more general statement.
                $endgroup$
                – L3435
                Jan 27 at 20:43
















              $begingroup$
              There is an elementary proof?
              $endgroup$
              – Problemsolving
              Jan 26 at 20:57




              $begingroup$
              There is an elementary proof?
              $endgroup$
              – Problemsolving
              Jan 26 at 20:57












              $begingroup$
              There are some more elementary proofs (like the ones below), but this one proves a more general statement.
              $endgroup$
              – L3435
              Jan 27 at 20:43




              $begingroup$
              There are some more elementary proofs (like the ones below), but this one proves a more general statement.
              $endgroup$
              – L3435
              Jan 27 at 20:43











              1












              $begingroup$

              Let $f(x)=a^x+b^x , f(3)=f(22)$



              By Holder inequality:



              $$ left( f(3) right )^{frac{1991}{2010}}cdot (f(2013))^{1-frac{1991}{2010}} ge f(22) Rightarrow f(2013)ge f(22)=f(3)$$



              $$left( f(22 right )^{frac{1}{1992}}cdot (f(2014))^{1-frac{1}{1992}} ge f(2013) Rightarrow f(2014) ge f(2013)$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $f(x)=a^x+b^x , f(3)=f(22)$



                By Holder inequality:



                $$ left( f(3) right )^{frac{1991}{2010}}cdot (f(2013))^{1-frac{1991}{2010}} ge f(22) Rightarrow f(2013)ge f(22)=f(3)$$



                $$left( f(22 right )^{frac{1}{1992}}cdot (f(2014))^{1-frac{1}{1992}} ge f(2013) Rightarrow f(2014) ge f(2013)$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $f(x)=a^x+b^x , f(3)=f(22)$



                  By Holder inequality:



                  $$ left( f(3) right )^{frac{1991}{2010}}cdot (f(2013))^{1-frac{1991}{2010}} ge f(22) Rightarrow f(2013)ge f(22)=f(3)$$



                  $$left( f(22 right )^{frac{1}{1992}}cdot (f(2014))^{1-frac{1}{1992}} ge f(2013) Rightarrow f(2014) ge f(2013)$$






                  share|cite|improve this answer









                  $endgroup$



                  Let $f(x)=a^x+b^x , f(3)=f(22)$



                  By Holder inequality:



                  $$ left( f(3) right )^{frac{1991}{2010}}cdot (f(2013))^{1-frac{1991}{2010}} ge f(22) Rightarrow f(2013)ge f(22)=f(3)$$



                  $$left( f(22 right )^{frac{1}{1992}}cdot (f(2014))^{1-frac{1}{1992}} ge f(2013) Rightarrow f(2014) ge f(2013)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 26 at 21:14









                  Sergic PrimazonSergic Primazon

                  53225




                  53225























                      0












                      $begingroup$

                      We need to prove that
                      $$sum_{cyc}left(a^{2014}-a^{2013}right)geq0$$ or
                      $$sum_{cyc}left(a^{2014}-a^{2013}-frac{1}{19}left(a^{22}-a^3right)right)geq0$$ or
                      $$sum_{cyc}a^3left(19a^{2011}-19a^{2010}-a^{19}+1right)geq0.$$
                      Now, by AM-GM
                      $$2010a^{2011}-2011a^{2010}+1geq0$$ or
                      $$19a^{2011}-frac{19cdot2011}{2010}a^{2010}+frac{19}{2010}geq0.$$
                      Id est, it's enough to prove that
                      $$left(frac{19cdot2011}{2010}-19right)a^{2010}-a^{19}+1-frac{19}{2010}geq0$$ or
                      $$19a^{2010}-2010a^{19}+1991geq0,$$ which is AM-GM again.



                      Done!






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        We need to prove that
                        $$sum_{cyc}left(a^{2014}-a^{2013}right)geq0$$ or
                        $$sum_{cyc}left(a^{2014}-a^{2013}-frac{1}{19}left(a^{22}-a^3right)right)geq0$$ or
                        $$sum_{cyc}a^3left(19a^{2011}-19a^{2010}-a^{19}+1right)geq0.$$
                        Now, by AM-GM
                        $$2010a^{2011}-2011a^{2010}+1geq0$$ or
                        $$19a^{2011}-frac{19cdot2011}{2010}a^{2010}+frac{19}{2010}geq0.$$
                        Id est, it's enough to prove that
                        $$left(frac{19cdot2011}{2010}-19right)a^{2010}-a^{19}+1-frac{19}{2010}geq0$$ or
                        $$19a^{2010}-2010a^{19}+1991geq0,$$ which is AM-GM again.



                        Done!






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          We need to prove that
                          $$sum_{cyc}left(a^{2014}-a^{2013}right)geq0$$ or
                          $$sum_{cyc}left(a^{2014}-a^{2013}-frac{1}{19}left(a^{22}-a^3right)right)geq0$$ or
                          $$sum_{cyc}a^3left(19a^{2011}-19a^{2010}-a^{19}+1right)geq0.$$
                          Now, by AM-GM
                          $$2010a^{2011}-2011a^{2010}+1geq0$$ or
                          $$19a^{2011}-frac{19cdot2011}{2010}a^{2010}+frac{19}{2010}geq0.$$
                          Id est, it's enough to prove that
                          $$left(frac{19cdot2011}{2010}-19right)a^{2010}-a^{19}+1-frac{19}{2010}geq0$$ or
                          $$19a^{2010}-2010a^{19}+1991geq0,$$ which is AM-GM again.



                          Done!






                          share|cite|improve this answer









                          $endgroup$



                          We need to prove that
                          $$sum_{cyc}left(a^{2014}-a^{2013}right)geq0$$ or
                          $$sum_{cyc}left(a^{2014}-a^{2013}-frac{1}{19}left(a^{22}-a^3right)right)geq0$$ or
                          $$sum_{cyc}a^3left(19a^{2011}-19a^{2010}-a^{19}+1right)geq0.$$
                          Now, by AM-GM
                          $$2010a^{2011}-2011a^{2010}+1geq0$$ or
                          $$19a^{2011}-frac{19cdot2011}{2010}a^{2010}+frac{19}{2010}geq0.$$
                          Id est, it's enough to prove that
                          $$left(frac{19cdot2011}{2010}-19right)a^{2010}-a^{19}+1-frac{19}{2010}geq0$$ or
                          $$19a^{2010}-2010a^{19}+1991geq0,$$ which is AM-GM again.



                          Done!







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 26 at 21:23









                          Michael RozenbergMichael Rozenberg

                          108k1895200




                          108k1895200






























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