How do I prove this relationship between positive terms of a G.P.?












2












$begingroup$



$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:



$$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$




This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.



$$A = frac1{ab}$$



$$B = frac1{cd}$$



$$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$



This gives the correct term on the left side as well as the $2$ on the right side.



But I am completely blank from here. How do I proceed from here to prove this relationship using means?










share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    $a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:



    $$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$




    This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.



    $$A = frac1{ab}$$



    $$B = frac1{cd}$$



    $$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$



    This gives the correct term on the left side as well as the $2$ on the right side.



    But I am completely blank from here. How do I proceed from here to prove this relationship using means?










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      $a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:



      $$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$




      This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.



      $$A = frac1{ab}$$



      $$B = frac1{cd}$$



      $$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$



      This gives the correct term on the left side as well as the $2$ on the right side.



      But I am completely blank from here. How do I proceed from here to prove this relationship using means?










      share|cite|improve this question











      $endgroup$





      $a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:



      $$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$




      This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.



      $$A = frac1{ab}$$



      $$B = frac1{cd}$$



      $$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$



      This gives the correct term on the left side as well as the $2$ on the right side.



      But I am completely blank from here. How do I proceed from here to prove this relationship using means?







      sequences-and-series inequality arithmetic-progressions geometric-series geometric-progressions






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 29 at 14:47









      Maria Mazur

      49.3k1360123




      49.3k1360123










      asked Jan 29 at 13:41









      user638500user638500

      403




      403






















          1 Answer
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          $begingroup$

          So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$



          or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$



          or $$boxed{x^4+1>2(x+x^3-x^2)}$$



          or



          $$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.





          You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$



          so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, I am terrible at maths. Can you please hint on how to proceed further?
            $endgroup$
            – user638500
            Jan 29 at 14:07












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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$



          or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$



          or $$boxed{x^4+1>2(x+x^3-x^2)}$$



          or



          $$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.





          You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$



          so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, I am terrible at maths. Can you please hint on how to proceed further?
            $endgroup$
            – user638500
            Jan 29 at 14:07
















          3












          $begingroup$

          So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$



          or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$



          or $$boxed{x^4+1>2(x+x^3-x^2)}$$



          or



          $$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.





          You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$



          so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, I am terrible at maths. Can you please hint on how to proceed further?
            $endgroup$
            – user638500
            Jan 29 at 14:07














          3












          3








          3





          $begingroup$

          So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$



          or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$



          or $$boxed{x^4+1>2(x+x^3-x^2)}$$



          or



          $$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.





          You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$



          so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.






          share|cite|improve this answer











          $endgroup$



          So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$



          or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$



          or $$boxed{x^4+1>2(x+x^3-x^2)}$$



          or



          $$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.





          You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$



          so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 29 at 14:45









          J. W. Tanner

          4,1961320




          4,1961320










          answered Jan 29 at 13:52









          Maria MazurMaria Mazur

          49.3k1360123




          49.3k1360123












          • $begingroup$
            Sorry, I am terrible at maths. Can you please hint on how to proceed further?
            $endgroup$
            – user638500
            Jan 29 at 14:07


















          • $begingroup$
            Sorry, I am terrible at maths. Can you please hint on how to proceed further?
            $endgroup$
            – user638500
            Jan 29 at 14:07
















          $begingroup$
          Sorry, I am terrible at maths. Can you please hint on how to proceed further?
          $endgroup$
          – user638500
          Jan 29 at 14:07




          $begingroup$
          Sorry, I am terrible at maths. Can you please hint on how to proceed further?
          $endgroup$
          – user638500
          Jan 29 at 14:07


















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