Prove: $log_2(x)+log_3(x)+log_5(x)>9log_30(x)$

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Prove for all $x>1$



$log_2(x)+log_3(x)+log_5(x)>9log_30(x)$




So what I did was:
beginalign
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(30)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(2cdot3cdot5)=9fracln(x)ln(2)+ln(3)+ln(5)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)ge3frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&!ln(2)+ln(3)+ln(5)ge3sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)>9fracln(x)ln(2)+ln(3)+ln(5)
endalign



I think this should prove the inequality but I'm not really sure how to formulate it properly if it was in an exam, any tips how to conclude? (if my proof is correct)




proof-verification inequality logarithms a.m.-g.m.-inequality




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  • 1




    Your proof is upside down, as you should start with something you know to be true, such as the arithmetic-geometric mean inequality, and conclude with what you are trying to prove. You should also probably use $x gt 1 implies ln(x) >0$ in your proof
    – Henry
    Jul 20 at 13:20











  • Probably write the symbol for the biconditional $$iff$$ in-between lines, to indicate that the logical flow goes both ways?
    – Jose Arnaldo Bebita Dris
    Jul 20 at 13:25














up vote
1
down vote

favorite













Prove for all $x>1$



$log_2(x)+log_3(x)+log_5(x)>9log_30(x)$




So what I did was:
beginalign
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(30)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(2cdot3cdot5)=9fracln(x)ln(2)+ln(3)+ln(5)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)ge3frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&!ln(2)+ln(3)+ln(5)ge3sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)>9fracln(x)ln(2)+ln(3)+ln(5)
endalign



I think this should prove the inequality but I'm not really sure how to formulate it properly if it was in an exam, any tips how to conclude? (if my proof is correct)




proof-verification inequality logarithms a.m.-g.m.-inequality




share|cite|improve this question

















  • 1




    Your proof is upside down, as you should start with something you know to be true, such as the arithmetic-geometric mean inequality, and conclude with what you are trying to prove. You should also probably use $x gt 1 implies ln(x) >0$ in your proof
    – Henry
    Jul 20 at 13:20











  • Probably write the symbol for the biconditional $$iff$$ in-between lines, to indicate that the logical flow goes both ways?
    – Jose Arnaldo Bebita Dris
    Jul 20 at 13:25












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Prove for all $x>1$



$log_2(x)+log_3(x)+log_5(x)>9log_30(x)$




So what I did was:
beginalign
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(30)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(2cdot3cdot5)=9fracln(x)ln(2)+ln(3)+ln(5)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)ge3frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&!ln(2)+ln(3)+ln(5)ge3sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)>9fracln(x)ln(2)+ln(3)+ln(5)
endalign



I think this should prove the inequality but I'm not really sure how to formulate it properly if it was in an exam, any tips how to conclude? (if my proof is correct)




proof-verification inequality logarithms a.m.-g.m.-inequality




share|cite|improve this question














Prove for all $x>1$



$log_2(x)+log_3(x)+log_5(x)>9log_30(x)$




So what I did was:
beginalign
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(30)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)>9fracln(x)ln(2cdot3cdot5)=9fracln(x)ln(2)+ln(3)+ln(5)
\[6px]
&fracln(x)ln(2)+fracln(x)ln(3)+fracln(x)ln(5)ge3frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&!ln(2)+ln(3)+ln(5)ge3sqrt[3]ln(2)ln(3)ln(5)
\[6px]
&frac3ln(x)sqrt[3]ln(2)ln(3)ln(5)>9fracln(x)ln(2)+ln(3)+ln(5)
endalign



I think this should prove the inequality but I'm not really sure how to formulate it properly if it was in an exam, any tips how to conclude? (if my proof is correct)





proof-verification inequality logarithms a.m.-g.m.-inequality





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edited Jul 20 at 13:50









egreg

164k1180187




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asked Jul 20 at 13:15









Maxim

616314




616314







  • 1




    Your proof is upside down, as you should start with something you know to be true, such as the arithmetic-geometric mean inequality, and conclude with what you are trying to prove. You should also probably use $x gt 1 implies ln(x) >0$ in your proof
    – Henry
    Jul 20 at 13:20











  • Probably write the symbol for the biconditional $$iff$$ in-between lines, to indicate that the logical flow goes both ways?
    – Jose Arnaldo Bebita Dris
    Jul 20 at 13:25












  • 1




    Your proof is upside down, as you should start with something you know to be true, such as the arithmetic-geometric mean inequality, and conclude with what you are trying to prove. You should also probably use $x gt 1 implies ln(x) >0$ in your proof
    – Henry
    Jul 20 at 13:20











  • Probably write the symbol for the biconditional $$iff$$ in-between lines, to indicate that the logical flow goes both ways?
    – Jose Arnaldo Bebita Dris
    Jul 20 at 13:25







1




1




Your proof is upside down, as you should start with something you know to be true, such as the arithmetic-geometric mean inequality, and conclude with what you are trying to prove. You should also probably use $x gt 1 implies ln(x) >0$ in your proof
– Henry
Jul 20 at 13:20





Your proof is upside down, as you should start with something you know to be true, such as the arithmetic-geometric mean inequality, and conclude with what you are trying to prove. You should also probably use $x gt 1 implies ln(x) >0$ in your proof
– Henry
Jul 20 at 13:20













Probably write the symbol for the biconditional $$iff$$ in-between lines, to indicate that the logical flow goes both ways?
– Jose Arnaldo Bebita Dris
Jul 20 at 13:25




Probably write the symbol for the biconditional $$iff$$ in-between lines, to indicate that the logical flow goes both ways?
– Jose Arnaldo Bebita Dris
Jul 20 at 13:25










3 Answers
3






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oldest

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up vote
3
down vote



accepted










Consider the left hand side:



$$log_2x+log_3x+log_5x = fraclog_30xlog_302+fraclog_30xlog_303+fraclog_30xlog_305.$$



Factor out $log_30x$:



$$log_2x+log_3x+log_5x = log_30xleft(frac1log_302+frac1log_305+frac1log_305right).$$



Now use the fact that $frac1log_ba = log_ab$:



$$log_2x+log_3x+log_5x = log_30xleft(log_230+log_330+log_530right).$$



So now all we need to show is



$$log_230+log_330+log_530>9.$$



$log_216=4$, and since $log_2$ is a strictly increasing function, this means $log_230>(log_216=)4$. Applying the same logic to the other two $log$s, we get that $log_330>3$ and $log_530>2$, hence



$$log_230+log_330+log_530>4+3+2=9$$



and we are done






share|cite|improve this answer






























    up vote
    2
    down vote













    From here since $ln x>0$



    $$fracln xln 2+fracln xln 3+fracln xln 5>9fracln xln 30iff frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$



    we can conclude, indeed by HM-AM inequality we have



    $$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
    implies
    frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$






    share|cite|improve this answer























    • Looks very interesting, never heard of a harmonic average before. 1.Is the harmonic average necessarily greater than the arithmetic average and not equal to it? 2.Could you please elaborate a bit more on the last step you did? I'm not sure how the HM-AM proves it completely.
      – Maxim
      Jul 20 at 14:20











    • @Maxim Harmonic mean is less or equal to AM and equality holds only if the positive terms are equals, for the proof and discussion refer to artofproblemsolving.com/wiki/…
      – gimusi
      Jul 20 at 14:21


















    up vote
    1
    down vote













    $$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$



    $$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$



    $$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$






    share|cite|improve this answer























    • aha but how did you do that last calculation? Because in an exam, I won't have a calculator to help me with that..
      – Maxim
      Jul 20 at 14:11







    • 1




      @Maxim OK, check my edit if you do not want to use calculator.
      – Mohammad Riazi-Kermani
      Jul 20 at 14:34










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    3 Answers
    3






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    3 Answers
    3






    active

    oldest

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    active

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    up vote
    3
    down vote



    accepted










    Consider the left hand side:



    $$log_2x+log_3x+log_5x = fraclog_30xlog_302+fraclog_30xlog_303+fraclog_30xlog_305.$$



    Factor out $log_30x$:



    $$log_2x+log_3x+log_5x = log_30xleft(frac1log_302+frac1log_305+frac1log_305right).$$



    Now use the fact that $frac1log_ba = log_ab$:



    $$log_2x+log_3x+log_5x = log_30xleft(log_230+log_330+log_530right).$$



    So now all we need to show is



    $$log_230+log_330+log_530>9.$$



    $log_216=4$, and since $log_2$ is a strictly increasing function, this means $log_230>(log_216=)4$. Applying the same logic to the other two $log$s, we get that $log_330>3$ and $log_530>2$, hence



    $$log_230+log_330+log_530>4+3+2=9$$



    and we are done






    share|cite|improve this answer



























      up vote
      3
      down vote



      accepted










      Consider the left hand side:



      $$log_2x+log_3x+log_5x = fraclog_30xlog_302+fraclog_30xlog_303+fraclog_30xlog_305.$$



      Factor out $log_30x$:



      $$log_2x+log_3x+log_5x = log_30xleft(frac1log_302+frac1log_305+frac1log_305right).$$



      Now use the fact that $frac1log_ba = log_ab$:



      $$log_2x+log_3x+log_5x = log_30xleft(log_230+log_330+log_530right).$$



      So now all we need to show is



      $$log_230+log_330+log_530>9.$$



      $log_216=4$, and since $log_2$ is a strictly increasing function, this means $log_230>(log_216=)4$. Applying the same logic to the other two $log$s, we get that $log_330>3$ and $log_530>2$, hence



      $$log_230+log_330+log_530>4+3+2=9$$



      and we are done






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Consider the left hand side:



        $$log_2x+log_3x+log_5x = fraclog_30xlog_302+fraclog_30xlog_303+fraclog_30xlog_305.$$



        Factor out $log_30x$:



        $$log_2x+log_3x+log_5x = log_30xleft(frac1log_302+frac1log_305+frac1log_305right).$$



        Now use the fact that $frac1log_ba = log_ab$:



        $$log_2x+log_3x+log_5x = log_30xleft(log_230+log_330+log_530right).$$



        So now all we need to show is



        $$log_230+log_330+log_530>9.$$



        $log_216=4$, and since $log_2$ is a strictly increasing function, this means $log_230>(log_216=)4$. Applying the same logic to the other two $log$s, we get that $log_330>3$ and $log_530>2$, hence



        $$log_230+log_330+log_530>4+3+2=9$$



        and we are done






        share|cite|improve this answer















        Consider the left hand side:



        $$log_2x+log_3x+log_5x = fraclog_30xlog_302+fraclog_30xlog_303+fraclog_30xlog_305.$$



        Factor out $log_30x$:



        $$log_2x+log_3x+log_5x = log_30xleft(frac1log_302+frac1log_305+frac1log_305right).$$



        Now use the fact that $frac1log_ba = log_ab$:



        $$log_2x+log_3x+log_5x = log_30xleft(log_230+log_330+log_530right).$$



        So now all we need to show is



        $$log_230+log_330+log_530>9.$$



        $log_216=4$, and since $log_2$ is a strictly increasing function, this means $log_230>(log_216=)4$. Applying the same logic to the other two $log$s, we get that $log_330>3$ and $log_530>2$, hence



        $$log_230+log_330+log_530>4+3+2=9$$



        and we are done







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 20 at 13:56


























        answered Jul 20 at 13:36









        aidangallagher4

        6191312




        6191312




















            up vote
            2
            down vote













            From here since $ln x>0$



            $$fracln xln 2+fracln xln 3+fracln xln 5>9fracln xln 30iff frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$



            we can conclude, indeed by HM-AM inequality we have



            $$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
            implies
            frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$






            share|cite|improve this answer























            • Looks very interesting, never heard of a harmonic average before. 1.Is the harmonic average necessarily greater than the arithmetic average and not equal to it? 2.Could you please elaborate a bit more on the last step you did? I'm not sure how the HM-AM proves it completely.
              – Maxim
              Jul 20 at 14:20











            • @Maxim Harmonic mean is less or equal to AM and equality holds only if the positive terms are equals, for the proof and discussion refer to artofproblemsolving.com/wiki/…
              – gimusi
              Jul 20 at 14:21















            up vote
            2
            down vote













            From here since $ln x>0$



            $$fracln xln 2+fracln xln 3+fracln xln 5>9fracln xln 30iff frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$



            we can conclude, indeed by HM-AM inequality we have



            $$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
            implies
            frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$






            share|cite|improve this answer























            • Looks very interesting, never heard of a harmonic average before. 1.Is the harmonic average necessarily greater than the arithmetic average and not equal to it? 2.Could you please elaborate a bit more on the last step you did? I'm not sure how the HM-AM proves it completely.
              – Maxim
              Jul 20 at 14:20











            • @Maxim Harmonic mean is less or equal to AM and equality holds only if the positive terms are equals, for the proof and discussion refer to artofproblemsolving.com/wiki/…
              – gimusi
              Jul 20 at 14:21













            up vote
            2
            down vote










            up vote
            2
            down vote









            From here since $ln x>0$



            $$fracln xln 2+fracln xln 3+fracln xln 5>9fracln xln 30iff frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$



            we can conclude, indeed by HM-AM inequality we have



            $$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
            implies
            frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$






            share|cite|improve this answer















            From here since $ln x>0$



            $$fracln xln 2+fracln xln 3+fracln xln 5>9fracln xln 30iff frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$



            we can conclude, indeed by HM-AM inequality we have



            $$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
            implies
            frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 20 at 13:49


























            answered Jul 20 at 13:44









            gimusi

            65.4k73584




            65.4k73584











            • Looks very interesting, never heard of a harmonic average before. 1.Is the harmonic average necessarily greater than the arithmetic average and not equal to it? 2.Could you please elaborate a bit more on the last step you did? I'm not sure how the HM-AM proves it completely.
              – Maxim
              Jul 20 at 14:20











            • @Maxim Harmonic mean is less or equal to AM and equality holds only if the positive terms are equals, for the proof and discussion refer to artofproblemsolving.com/wiki/…
              – gimusi
              Jul 20 at 14:21

















            • Looks very interesting, never heard of a harmonic average before. 1.Is the harmonic average necessarily greater than the arithmetic average and not equal to it? 2.Could you please elaborate a bit more on the last step you did? I'm not sure how the HM-AM proves it completely.
              – Maxim
              Jul 20 at 14:20











            • @Maxim Harmonic mean is less or equal to AM and equality holds only if the positive terms are equals, for the proof and discussion refer to artofproblemsolving.com/wiki/…
              – gimusi
              Jul 20 at 14:21
















            Looks very interesting, never heard of a harmonic average before. 1.Is the harmonic average necessarily greater than the arithmetic average and not equal to it? 2.Could you please elaborate a bit more on the last step you did? I'm not sure how the HM-AM proves it completely.
            – Maxim
            Jul 20 at 14:20





            Looks very interesting, never heard of a harmonic average before. 1.Is the harmonic average necessarily greater than the arithmetic average and not equal to it? 2.Could you please elaborate a bit more on the last step you did? I'm not sure how the HM-AM proves it completely.
            – Maxim
            Jul 20 at 14:20













            @Maxim Harmonic mean is less or equal to AM and equality holds only if the positive terms are equals, for the proof and discussion refer to artofproblemsolving.com/wiki/…
            – gimusi
            Jul 20 at 14:21





            @Maxim Harmonic mean is less or equal to AM and equality holds only if the positive terms are equals, for the proof and discussion refer to artofproblemsolving.com/wiki/…
            – gimusi
            Jul 20 at 14:21











            up vote
            1
            down vote













            $$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$



            $$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$



            $$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$






            share|cite|improve this answer























            • aha but how did you do that last calculation? Because in an exam, I won't have a calculator to help me with that..
              – Maxim
              Jul 20 at 14:11







            • 1




              @Maxim OK, check my edit if you do not want to use calculator.
              – Mohammad Riazi-Kermani
              Jul 20 at 14:34














            up vote
            1
            down vote













            $$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$



            $$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$



            $$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$






            share|cite|improve this answer























            • aha but how did you do that last calculation? Because in an exam, I won't have a calculator to help me with that..
              – Maxim
              Jul 20 at 14:11







            • 1




              @Maxim OK, check my edit if you do not want to use calculator.
              – Mohammad Riazi-Kermani
              Jul 20 at 14:34












            up vote
            1
            down vote










            up vote
            1
            down vote









            $$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$



            $$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$



            $$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$






            share|cite|improve this answer















            $$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$



            $$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$



            $$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 20 at 14:33


























            answered Jul 20 at 14:04









            Mohammad Riazi-Kermani

            27.5k41852




            27.5k41852











            • aha but how did you do that last calculation? Because in an exam, I won't have a calculator to help me with that..
              – Maxim
              Jul 20 at 14:11







            • 1




              @Maxim OK, check my edit if you do not want to use calculator.
              – Mohammad Riazi-Kermani
              Jul 20 at 14:34
















            • aha but how did you do that last calculation? Because in an exam, I won't have a calculator to help me with that..
              – Maxim
              Jul 20 at 14:11







            • 1




              @Maxim OK, check my edit if you do not want to use calculator.
              – Mohammad Riazi-Kermani
              Jul 20 at 14:34















            aha but how did you do that last calculation? Because in an exam, I won't have a calculator to help me with that..
            – Maxim
            Jul 20 at 14:11





            aha but how did you do that last calculation? Because in an exam, I won't have a calculator to help me with that..
            – Maxim
            Jul 20 at 14:11





            1




            1




            @Maxim OK, check my edit if you do not want to use calculator.
            – Mohammad Riazi-Kermani
            Jul 20 at 14:34




            @Maxim OK, check my edit if you do not want to use calculator.
            – Mohammad Riazi-Kermani
            Jul 20 at 14:34












             

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