chebyshev's inequality - Question












4














I had a question in my exam and they asked to prove that
prove that:




$$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.




Now , I solved it , but the problem is that in the answer they wrote this:
using Chebyshev inequality:



$$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
can someone please send me link regarding this inequality or just write it here?



Thank you.






algebra-precalculus inequality symmetric-polynomials rearrangement-inequality







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    4














    I had a question in my exam and they asked to prove that
    prove that:




    $$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.




    Now , I solved it , but the problem is that in the answer they wrote this:
    using Chebyshev inequality:



    $$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
    And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
    can someone please send me link regarding this inequality or just write it here?



    Thank you.






    algebra-precalculus inequality symmetric-polynomials rearrangement-inequality







    share|cite|improve this question



























      4












      4








      4







      I had a question in my exam and they asked to prove that
      prove that:




      $$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.




      Now , I solved it , but the problem is that in the answer they wrote this:
      using Chebyshev inequality:



      $$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
      And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
      can someone please send me link regarding this inequality or just write it here?



      Thank you.






      algebra-precalculus inequality symmetric-polynomials rearrangement-inequality







      share|cite|improve this question















      I had a question in my exam and they asked to prove that
      prove that:




      $$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.




      Now , I solved it , but the problem is that in the answer they wrote this:
      using Chebyshev inequality:



      $$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
      And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
      can someone please send me link regarding this inequality or just write it here?



      Thank you.






      algebra-precalculus inequality symmetric-polynomials rearrangement-inequality




      algebra-precalculus inequality symmetric-polynomials rearrangement-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 1:24









      Michael Rozenberg

      97.5k1589188




      97.5k1589188










      asked Dec 31 '18 at 23:59









      orel Zemachorel Zemach

      232




      232






















          1 Answer
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          1














          It's just C-S:
          $$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$



          The Chebyshov inequality it's the following.



          Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.



          Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
          The proof of this inequality follows immediately from Rearrangement.



          There is also the following way:



          $$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
          Id est, it's enough to prove that:
          $$3(1-a+a^2)geq1+a+a^2$$ or
          $$(a-1)^2geq0.$$






          share|cite|improve this answer























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            1 Answer
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            active

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            1














            It's just C-S:
            $$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$



            The Chebyshov inequality it's the following.



            Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.



            Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
            The proof of this inequality follows immediately from Rearrangement.



            There is also the following way:



            $$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
            Id est, it's enough to prove that:
            $$3(1-a+a^2)geq1+a+a^2$$ or
            $$(a-1)^2geq0.$$






            share|cite|improve this answer




























              1














              It's just C-S:
              $$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$



              The Chebyshov inequality it's the following.



              Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.



              Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
              The proof of this inequality follows immediately from Rearrangement.



              There is also the following way:



              $$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
              Id est, it's enough to prove that:
              $$3(1-a+a^2)geq1+a+a^2$$ or
              $$(a-1)^2geq0.$$






              share|cite|improve this answer


























                1












                1








                1






                It's just C-S:
                $$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$



                The Chebyshov inequality it's the following.



                Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.



                Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
                The proof of this inequality follows immediately from Rearrangement.



                There is also the following way:



                $$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
                Id est, it's enough to prove that:
                $$3(1-a+a^2)geq1+a+a^2$$ or
                $$(a-1)^2geq0.$$






                share|cite|improve this answer














                It's just C-S:
                $$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$



                The Chebyshov inequality it's the following.



                Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.



                Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
                The proof of this inequality follows immediately from Rearrangement.



                There is also the following way:



                $$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
                Id est, it's enough to prove that:
                $$3(1-a+a^2)geq1+a+a^2$$ or
                $$(a-1)^2geq0.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 1 at 1:20

























                answered Jan 1 at 0:48









                Michael RozenbergMichael Rozenberg

                97.5k1589188




                97.5k1589188






























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