Duplicate - Proof by Ordinary Induction: $a^n-b^n leq na^{n-1}(a-b)$












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This question has already been asked:



Proving Inequality using Induction $a^n-b^n leq na^{n-1}(a-b)$



However has not been answered properly. (Even thought the OP checked an answer)



The answers provided are direct proofs and not induction proofs.



Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:



$$a^n-b^n leq na^{n-1}(a-b)$$



Anybody has ideas?






inequality induction







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$endgroup$

















    0












    $begingroup$


    This question has already been asked:



    Proving Inequality using Induction $a^n-b^n leq na^{n-1}(a-b)$



    However has not been answered properly. (Even thought the OP checked an answer)



    The answers provided are direct proofs and not induction proofs.



    Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:



    $$a^n-b^n leq na^{n-1}(a-b)$$



    Anybody has ideas?






    inequality induction







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      This question has already been asked:



      Proving Inequality using Induction $a^n-b^n leq na^{n-1}(a-b)$



      However has not been answered properly. (Even thought the OP checked an answer)



      The answers provided are direct proofs and not induction proofs.



      Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:



      $$a^n-b^n leq na^{n-1}(a-b)$$



      Anybody has ideas?






      inequality induction







      share|cite|improve this question











      $endgroup$




      This question has already been asked:



      Proving Inequality using Induction $a^n-b^n leq na^{n-1}(a-b)$



      However has not been answered properly. (Even thought the OP checked an answer)



      The answers provided are direct proofs and not induction proofs.



      Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:



      $$a^n-b^n leq na^{n-1}(a-b)$$



      Anybody has ideas?






      inequality induction




      inequality induction






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 13 '17 at 12:20









      Community♦

      1




      1










      asked Mar 6 '16 at 21:08









      ex.nihilex.nihil

      220111




      220111






















          2 Answers
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          active

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          3












          $begingroup$

          We do the induction step. Suppose that for a certain $k$ we have $a^k-b^kle ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}le (k+1)a^k(b-a)$.



          We have
          $$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).tag{1}$$
          Note that by the induction hypothesis $b(a^k-b^k)le a(a^k-b^k)le ka^k(a-b)$. So the sum on the right-hand side of (1) is $le (k+1)a^k(a-b)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas.
            $endgroup$
            – ex.nihil
            Mar 6 '16 at 21:34





















          -1












          $begingroup$

          The proof proceeds by induction. Here we will need to make use of two base
          cases, since it turns out that n = 1 needs to be treated separately.



          (1) First base case (n = 1): here a^1 −b^1 ≤ 1*a^0(a−b) = a − b,


          so the statement holds.



          (2) Second base case (n = 2):
          
a^2 − b^2 = (a + b)(a − b)
          
< 2a(a − b)    since a > b
          
= na^(n−1)(a − b), for n = 2
          

          
(3) Inductive step: Now suppose that a^n − b^n < na^(n−1)(a − b), where 0 < 
          
b < a. Then
          
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
          
< a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
          
< a^n(a − b) + ana^(n−1)(a − b) since b < a
          
= (n + 1)a^n(a − b)
          
so that a^(n+1) − b^(n+1) < (n + 1)a^n(a − b) for n > 2.





          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            We do the induction step. Suppose that for a certain $k$ we have $a^k-b^kle ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}le (k+1)a^k(b-a)$.



            We have
            $$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).tag{1}$$
            Note that by the induction hypothesis $b(a^k-b^k)le a(a^k-b^k)le ka^k(a-b)$. So the sum on the right-hand side of (1) is $le (k+1)a^k(a-b)$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas.
              $endgroup$
              – ex.nihil
              Mar 6 '16 at 21:34


















            3












            $begingroup$

            We do the induction step. Suppose that for a certain $k$ we have $a^k-b^kle ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}le (k+1)a^k(b-a)$.



            We have
            $$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).tag{1}$$
            Note that by the induction hypothesis $b(a^k-b^k)le a(a^k-b^k)le ka^k(a-b)$. So the sum on the right-hand side of (1) is $le (k+1)a^k(a-b)$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas.
              $endgroup$
              – ex.nihil
              Mar 6 '16 at 21:34
















            3












            3








            3





            $begingroup$

            We do the induction step. Suppose that for a certain $k$ we have $a^k-b^kle ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}le (k+1)a^k(b-a)$.



            We have
            $$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).tag{1}$$
            Note that by the induction hypothesis $b(a^k-b^k)le a(a^k-b^k)le ka^k(a-b)$. So the sum on the right-hand side of (1) is $le (k+1)a^k(a-b)$.






            share|cite|improve this answer











            $endgroup$



            We do the induction step. Suppose that for a certain $k$ we have $a^k-b^kle ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}le (k+1)a^k(b-a)$.



            We have
            $$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).tag{1}$$
            Note that by the induction hypothesis $b(a^k-b^k)le a(a^k-b^k)le ka^k(a-b)$. So the sum on the right-hand side of (1) is $le (k+1)a^k(a-b)$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 6 '16 at 21:36

























            answered Mar 6 '16 at 21:23









            André NicolasAndré Nicolas

            452k36423808




            452k36423808








            • 1




              $begingroup$
              I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas.
              $endgroup$
              – ex.nihil
              Mar 6 '16 at 21:34
















            • 1




              $begingroup$
              I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas.
              $endgroup$
              – ex.nihil
              Mar 6 '16 at 21:34










            1




            1




            $begingroup$
            I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas.
            $endgroup$
            – ex.nihil
            Mar 6 '16 at 21:34






            $begingroup$
            I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas.
            $endgroup$
            – ex.nihil
            Mar 6 '16 at 21:34













            -1












            $begingroup$

            The proof proceeds by induction. Here we will need to make use of two base
            cases, since it turns out that n = 1 needs to be treated separately.



            (1) First base case (n = 1): here a^1 −b^1 ≤ 1*a^0(a−b) = a − b,


            so the statement holds.



            (2) Second base case (n = 2):
            
a^2 − b^2 = (a + b)(a − b)
            
< 2a(a − b)    since a > b
            
= na^(n−1)(a − b), for n = 2
            

            
(3) Inductive step: Now suppose that a^n − b^n < na^(n−1)(a − b), where 0 < 
            
b < a. Then
            
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
            
< a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
            
< a^n(a − b) + ana^(n−1)(a − b) since b < a
            
= (n + 1)a^n(a − b)
            
so that a^(n+1) − b^(n+1) < (n + 1)a^n(a − b) for n > 2.





            share|cite|improve this answer











            $endgroup$


















              -1












              $begingroup$

              The proof proceeds by induction. Here we will need to make use of two base
              cases, since it turns out that n = 1 needs to be treated separately.



              (1) First base case (n = 1): here a^1 −b^1 ≤ 1*a^0(a−b) = a − b,


              so the statement holds.



              (2) Second base case (n = 2):
              
a^2 − b^2 = (a + b)(a − b)
              
< 2a(a − b)    since a > b
              
= na^(n−1)(a − b), for n = 2
              

              
(3) Inductive step: Now suppose that a^n − b^n < na^(n−1)(a − b), where 0 < 
              
b < a. Then
              
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
              
< a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
              
< a^n(a − b) + ana^(n−1)(a − b) since b < a
              
= (n + 1)a^n(a − b)
              
so that a^(n+1) − b^(n+1) < (n + 1)a^n(a − b) for n > 2.





              share|cite|improve this answer











              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                The proof proceeds by induction. Here we will need to make use of two base
                cases, since it turns out that n = 1 needs to be treated separately.



                (1) First base case (n = 1): here a^1 −b^1 ≤ 1*a^0(a−b) = a − b,


                so the statement holds.



                (2) Second base case (n = 2):
                
a^2 − b^2 = (a + b)(a − b)
                
< 2a(a − b)    since a > b
                
= na^(n−1)(a − b), for n = 2
                

                
(3) Inductive step: Now suppose that a^n − b^n < na^(n−1)(a − b), where 0 < 
                
b < a. Then
                
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
                
< a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
                
< a^n(a − b) + ana^(n−1)(a − b) since b < a
                
= (n + 1)a^n(a − b)
                
so that a^(n+1) − b^(n+1) < (n + 1)a^n(a − b) for n > 2.





                share|cite|improve this answer











                $endgroup$



                The proof proceeds by induction. Here we will need to make use of two base
                cases, since it turns out that n = 1 needs to be treated separately.



                (1) First base case (n = 1): here a^1 −b^1 ≤ 1*a^0(a−b) = a − b,


                so the statement holds.



                (2) Second base case (n = 2):
                
a^2 − b^2 = (a + b)(a − b)
                
< 2a(a − b)    since a > b
                
= na^(n−1)(a − b), for n = 2
                

                
(3) Inductive step: Now suppose that a^n − b^n < na^(n−1)(a − b), where 0 < 
                
b < a. Then
                
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
                
< a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
                
< a^n(a − b) + ana^(n−1)(a − b) since b < a
                
= (n + 1)a^n(a − b)
                
so that a^(n+1) − b^(n+1) < (n + 1)a^n(a − b) for n > 2.






                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 7 '18 at 16:52

























                answered Dec 7 '18 at 16:30









                kiro malakkiro malak

                11




                11






























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