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Prove: $log_2(x)+log_3(x)+log_5(x)>9log_30(x)$
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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
edited Jul 20 at 13:50
egreg
164k1180187
asked Jul 20 at 13:15
Maxim
616314
add a comment |Â
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
edited Jul 20 at 13:50
egreg
164k1180187
asked Jul 20 at 13:15
Maxim
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
add a comment |Â
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
edited Jul 20 at 13:50
egreg
164k1180187
asked Jul 20 at 13:15
Maxim
616314
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
edited Jul 20 at 13:50
egreg
164k1180187
asked Jul 20 at 13:15
Maxim
616314
edited Jul 20 at 13:50
egreg
164k1180187
edited Jul 20 at 13:50
egreg
164k1180187
edited Jul 20 at 13:50
egreg
164k1180187
164k1180187
asked Jul 20 at 13:15
Maxim
616314
asked Jul 20 at 13:15
Maxim
616314
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
add a comment |Â
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
add a comment |Â
3 Answers
3
active
oldest
votes
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
answered Jul 20 at 13:36
aidangallagher4
6191312
add a comment |Â
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
$$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
implies
frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$
answered Jul 20 at 13:44
gimusi
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
add a comment |Â
up vote
1
down vote
$$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$
$$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$
$$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$
answered Jul 20 at 14:04
Mohammad Riazi-Kermani
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
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
answered Jul 20 at 13:36
aidangallagher4
6191312
add a comment |Â
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
answered Jul 20 at 13:36
aidangallagher4
6191312
add a comment |Â
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
answered Jul 20 at 13:36
aidangallagher4
6191312
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
answered Jul 20 at 13:36
aidangallagher4
6191312
edited Jul 20 at 13:56
answered Jul 20 at 13:36
aidangallagher4
6191312
answered Jul 20 at 13:36
aidangallagher4
6191312
answered Jul 20 at 13:36
aidangallagher4
6191312
6191312
add a comment |Â
add a comment |Â
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
$$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
implies
frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$
answered Jul 20 at 13:44
gimusi
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
add a comment |Â
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
$$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
implies
frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$
answered Jul 20 at 13:44
gimusi
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
add a comment |Â
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
$$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
implies
frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$
answered Jul 20 at 13:44
gimusi
65.4k73584
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
$$frac3frac1ln 2+frac1ln 3+frac1ln 5<fracln 2+ln 3+ln 53=fracln 303
implies
frac1ln 2+frac1ln 3+frac1ln 5>frac9ln 30$$
answered Jul 20 at 13:44
gimusi
65.4k73584
edited Jul 20 at 13:49
answered Jul 20 at 13:44
gimusi
65.4k73584
answered Jul 20 at 13:44
gimusi
65.4k73584
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
add a comment |Â
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
add a comment |Â
up vote
1
down vote
$$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$
$$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$
$$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$
answered Jul 20 at 14:04
Mohammad Riazi-Kermani
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
add a comment |Â
up vote
1
down vote
$$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$
$$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$
$$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$
answered Jul 20 at 14:04
Mohammad Riazi-Kermani
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
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$
$$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$
$$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$
answered Jul 20 at 14:04
Mohammad Riazi-Kermani
27.5k41852
$$log_2(x)+log_3(x)+log_5(x)>9log_30(x) iff frac log(x)log2 + frac log(x)log3+frac log(x)log5 >9 frac log(x)log30$$
$$iff frac 1log2 + frac 1log3+frac 1log5 >9 frac 1log30$$
$$iff log_2 30 +log_3 30 + log_5 30 > log_2 16 +log_3 27 + log_5 25 =4+3+2= 9 $$
answered Jul 20 at 14:04
Mohammad Riazi-Kermani
27.5k41852
edited Jul 20 at 14:33
answered Jul 20 at 14:04
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 14:04
Mohammad Riazi-Kermani
27.5k41852
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
add a comment |Â
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
add a comment |Â
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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