Mixing
Prove or disprove the inequality $ -xln (x)leq ln(1-x)(ln (x) + 1 - x)$
$begingroup$
how can one disprove or prove the equation below? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ -xln (x)leq ln(1-x)(ln (x) + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
Any help will be quite useful to me. Thanks.
inequality logarithms
edited Jan 11 at 18:06
6005
36.2k751125
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
$endgroup$
|
show 4 more comments
$begingroup$
how can one disprove or prove the equation below? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ -xln (x)leq ln(1-x)(ln (x) + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
Any help will be quite useful to me. Thanks.
inequality logarithms
edited Jan 11 at 18:06
6005
36.2k751125
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
$endgroup$
1
$begingroup$
prove/disprove, not "approve/disapprove"
$endgroup$
– rschwieb
Jan 11 at 17:49
$begingroup$
Do you mean $$-xln(x)le ln(1-x)(ln(x)+1-x)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:52
$begingroup$
@Dr.SonnhardGraubner yes I mean that
$endgroup$
– Weez Khan
Jan 11 at 17:53
$begingroup$
I think you mean $ge$, not $le$. The inequality you wrote down is not true for $0 < x < 1$, but if you switch it to $ge$ instead of $le$ then it is true.
$endgroup$
– 6005
Jan 11 at 17:55
$begingroup$
This inequalitiy is not true, take $$x=frac{1}{2}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:56
|
show 4 more comments
$begingroup$
how can one disprove or prove the equation below? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ -xln (x)leq ln(1-x)(ln (x) + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
Any help will be quite useful to me. Thanks.
inequality logarithms
edited Jan 11 at 18:06
6005
36.2k751125
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
$endgroup$
how can one disprove or prove the equation below? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ -xln (x)leq ln(1-x)(ln (x) + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
Any help will be quite useful to me. Thanks.
inequality logarithms
inequality logarithms
edited Jan 11 at 18:06
6005
36.2k751125
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
edited Jan 11 at 18:06
6005
36.2k751125
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
edited Jan 11 at 18:06
6005
36.2k751125
edited Jan 11 at 18:06
6005
36.2k751125
edited Jan 11 at 18:06
6005
36.2k751125
36.2k751125
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
asked Jan 11 at 17:46
Weez KhanWeez Khan
135
135
1
$begingroup$
prove/disprove, not "approve/disapprove"
$endgroup$
– rschwieb
Jan 11 at 17:49
$begingroup$
Do you mean $$-xln(x)le ln(1-x)(ln(x)+1-x)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:52
$begingroup$
@Dr.SonnhardGraubner yes I mean that
$endgroup$
– Weez Khan
Jan 11 at 17:53
$begingroup$
I think you mean $ge$, not $le$. The inequality you wrote down is not true for $0 < x < 1$, but if you switch it to $ge$ instead of $le$ then it is true.
$endgroup$
– 6005
Jan 11 at 17:55
$begingroup$
This inequalitiy is not true, take $$x=frac{1}{2}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:56
|
show 4 more comments
1
$begingroup$
prove/disprove, not "approve/disapprove"
$endgroup$
– rschwieb
Jan 11 at 17:49
$begingroup$
Do you mean $$-xln(x)le ln(1-x)(ln(x)+1-x)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:52
$begingroup$
@Dr.SonnhardGraubner yes I mean that
$endgroup$
– Weez Khan
Jan 11 at 17:53
$begingroup$
I think you mean $ge$, not $le$. The inequality you wrote down is not true for $0 < x < 1$, but if you switch it to $ge$ instead of $le$ then it is true.
$endgroup$
– 6005
Jan 11 at 17:55
$begingroup$
This inequalitiy is not true, take $$x=frac{1}{2}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:56
1
1
$begingroup$
prove/disprove, not "approve/disapprove"
$endgroup$
– rschwieb
Jan 11 at 17:49
$begingroup$
prove/disprove, not "approve/disapprove"
$endgroup$
– rschwieb
Jan 11 at 17:49
$begingroup$
Do you mean $$-xln(x)le ln(1-x)(ln(x)+1-x)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:52
$begingroup$
Do you mean $$-xln(x)le ln(1-x)(ln(x)+1-x)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:52
$begingroup$
@Dr.SonnhardGraubner yes I mean that
$endgroup$
– Weez Khan
Jan 11 at 17:53
$begingroup$
@Dr.SonnhardGraubner yes I mean that
$endgroup$
– Weez Khan
Jan 11 at 17:53
$begingroup$
I think you mean $ge$, not $le$. The inequality you wrote down is not true for $0 < x < 1$, but if you switch it to $ge$ instead of $le$ then it is true.
$endgroup$
– 6005
Jan 11 at 17:55
$begingroup$
I think you mean $ge$, not $le$. The inequality you wrote down is not true for $0 < x < 1$, but if you switch it to $ge$ instead of $le$ then it is true.
$endgroup$
– 6005
Jan 11 at 17:55
$begingroup$
This inequalitiy is not true, take $$x=frac{1}{2}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:56
$begingroup$
This inequalitiy is not true, take $$x=frac{1}{2}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:56
|
show 4 more comments
4 Answers
4
active
oldest
votes
$begingroup$
The inequality
$$ -xln xleq ln(1-x)(ln x + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
is not true.
For example, if we put in $x = frac{1}{2}$ we get
$$
- frac{1}{2} ln frac{1}{2} = frac12 ln 2 = 0.34ldots
$$
and
$$
lnleft(1 - frac12right)
left( ln frac12 + 1 - frac12right)
= left(- ln 2 right) left( frac12 - ln 2 right) = (ln 2)^2 - frac12 ln 2 = 0.13 ldots
$$
and $0.34 ldots > 0.13ldots$.
You probably made a mistake when you reduced it to this form. If you reverse the inequality, it would be true.
See the graph below:
We can see that the graph looks to be $> 0$ for $0 < x < 1$.
So we conjecture that the inequality
$$
- x ln x ge ln(1-x) (ln x + 1 - x)
$$
will be true.
answered Jan 11 at 18:04
60056005
36.2k751125
$endgroup$
$begingroup$
Thank you, just one more question, can we reduce the equation further(make it more simpler) so that it is intuitively possible as to why it is not possible
$endgroup$
– Weez Khan
Jan 11 at 18:06
add a comment |
$begingroup$
Hint: We have $$-ln(1-x)(ln(x)+1-x)-xln(x)geq 0$$ for $$0<x<1$$
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
Define
begin{align*}
f(x) = -xlog(x) - log(1-x)(log(x) + 1 - x)
end{align*}
It suffices to prove $f(x) ge 0$ for $x in [0, 1]$. Note that
begin{align*}
f(0) &= f(1) = 0 \
f'(x) &= frac{(x-1)log(1-x)}{x} + frac{xlog(x)}{1-x} begin{cases}
> 0 & x < frac{1}{2} \
= 0 & x = frac{1}{2} \
< 0 & x > frac{1}{2}
end{cases}
end{align*}
Therefore, $f(x) ge min(f(0), f(1)) = 0$ for $x in [0, 1]$, as desired.
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
$endgroup$
add a comment |
$begingroup$
Consider the function
$$f(x)=xln (x)+ ln(1-x),(ln (x) + 1 - x) $$ and notice the symmetry which makes that
$$f(x)+f(1-x)=0$$ Compute $f(x)$ anywhere to show that is negative. Using $x=frac 12$, you have $(log (2)-1) log (2)$ which is the minimum value.
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The inequality
$$ -xln xleq ln(1-x)(ln x + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
is not true.
For example, if we put in $x = frac{1}{2}$ we get
$$
- frac{1}{2} ln frac{1}{2} = frac12 ln 2 = 0.34ldots
$$
and
$$
lnleft(1 - frac12right)
left( ln frac12 + 1 - frac12right)
= left(- ln 2 right) left( frac12 - ln 2 right) = (ln 2)^2 - frac12 ln 2 = 0.13 ldots
$$
and $0.34 ldots > 0.13ldots$.
You probably made a mistake when you reduced it to this form. If you reverse the inequality, it would be true.
See the graph below:
We can see that the graph looks to be $> 0$ for $0 < x < 1$.
So we conjecture that the inequality
$$
- x ln x ge ln(1-x) (ln x + 1 - x)
$$
will be true.
answered Jan 11 at 18:04
60056005
36.2k751125
$endgroup$
$begingroup$
Thank you, just one more question, can we reduce the equation further(make it more simpler) so that it is intuitively possible as to why it is not possible
$endgroup$
– Weez Khan
Jan 11 at 18:06
add a comment |
$begingroup$
The inequality
$$ -xln xleq ln(1-x)(ln x + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
is not true.
For example, if we put in $x = frac{1}{2}$ we get
$$
- frac{1}{2} ln frac{1}{2} = frac12 ln 2 = 0.34ldots
$$
and
$$
lnleft(1 - frac12right)
left( ln frac12 + 1 - frac12right)
= left(- ln 2 right) left( frac12 - ln 2 right) = (ln 2)^2 - frac12 ln 2 = 0.13 ldots
$$
and $0.34 ldots > 0.13ldots$.
You probably made a mistake when you reduced it to this form. If you reverse the inequality, it would be true.
See the graph below:
We can see that the graph looks to be $> 0$ for $0 < x < 1$.
So we conjecture that the inequality
$$
- x ln x ge ln(1-x) (ln x + 1 - x)
$$
will be true.
answered Jan 11 at 18:04
60056005
36.2k751125
$endgroup$
$begingroup$
Thank you, just one more question, can we reduce the equation further(make it more simpler) so that it is intuitively possible as to why it is not possible
$endgroup$
– Weez Khan
Jan 11 at 18:06
add a comment |
$begingroup$
The inequality
$$ -xln xleq ln(1-x)(ln x + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
is not true.
For example, if we put in $x = frac{1}{2}$ we get
$$
- frac{1}{2} ln frac{1}{2} = frac12 ln 2 = 0.34ldots
$$
and
$$
lnleft(1 - frac12right)
left( ln frac12 + 1 - frac12right)
= left(- ln 2 right) left( frac12 - ln 2 right) = (ln 2)^2 - frac12 ln 2 = 0.13 ldots
$$
and $0.34 ldots > 0.13ldots$.
You probably made a mistake when you reduced it to this form. If you reverse the inequality, it would be true.
See the graph below:
We can see that the graph looks to be $> 0$ for $0 < x < 1$.
So we conjecture that the inequality
$$
- x ln x ge ln(1-x) (ln x + 1 - x)
$$
will be true.
answered Jan 11 at 18:04
60056005
36.2k751125
$endgroup$
The inequality
$$ -xln xleq ln(1-x)(ln x + 1 - x) qquadtext{for}qquad 0lt xlt 1$$
is not true.
For example, if we put in $x = frac{1}{2}$ we get
$$
- frac{1}{2} ln frac{1}{2} = frac12 ln 2 = 0.34ldots
$$
and
$$
lnleft(1 - frac12right)
left( ln frac12 + 1 - frac12right)
= left(- ln 2 right) left( frac12 - ln 2 right) = (ln 2)^2 - frac12 ln 2 = 0.13 ldots
$$
and $0.34 ldots > 0.13ldots$.
You probably made a mistake when you reduced it to this form. If you reverse the inequality, it would be true.
See the graph below:
We can see that the graph looks to be $> 0$ for $0 < x < 1$.
So we conjecture that the inequality
$$
- x ln x ge ln(1-x) (ln x + 1 - x)
$$
will be true.
answered Jan 11 at 18:04
60056005
36.2k751125
answered Jan 11 at 18:04
60056005
36.2k751125
answered Jan 11 at 18:04
60056005
36.2k751125
answered Jan 11 at 18:04
60056005
36.2k751125
36.2k751125
$begingroup$
Thank you, just one more question, can we reduce the equation further(make it more simpler) so that it is intuitively possible as to why it is not possible
$endgroup$
– Weez Khan
Jan 11 at 18:06
add a comment |
$begingroup$
Thank you, just one more question, can we reduce the equation further(make it more simpler) so that it is intuitively possible as to why it is not possible
$endgroup$
– Weez Khan
Jan 11 at 18:06
$begingroup$
Thank you, just one more question, can we reduce the equation further(make it more simpler) so that it is intuitively possible as to why it is not possible
$endgroup$
– Weez Khan
Jan 11 at 18:06
$begingroup$
Thank you, just one more question, can we reduce the equation further(make it more simpler) so that it is intuitively possible as to why it is not possible
$endgroup$
– Weez Khan
Jan 11 at 18:06
add a comment |
$begingroup$
Hint: We have $$-ln(1-x)(ln(x)+1-x)-xln(x)geq 0$$ for $$0<x<1$$
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
Hint: We have $$-ln(1-x)(ln(x)+1-x)-xln(x)geq 0$$ for $$0<x<1$$
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
Hint: We have $$-ln(1-x)(ln(x)+1-x)-xln(x)geq 0$$ for $$0<x<1$$
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
Hint: We have $$-ln(1-x)(ln(x)+1-x)-xln(x)geq 0$$ for $$0<x<1$$
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 11 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
75.1k42865
add a comment |
add a comment |
$begingroup$
Define
begin{align*}
f(x) = -xlog(x) - log(1-x)(log(x) + 1 - x)
end{align*}
It suffices to prove $f(x) ge 0$ for $x in [0, 1]$. Note that
begin{align*}
f(0) &= f(1) = 0 \
f'(x) &= frac{(x-1)log(1-x)}{x} + frac{xlog(x)}{1-x} begin{cases}
> 0 & x < frac{1}{2} \
= 0 & x = frac{1}{2} \
< 0 & x > frac{1}{2}
end{cases}
end{align*}
Therefore, $f(x) ge min(f(0), f(1)) = 0$ for $x in [0, 1]$, as desired.
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
$endgroup$
add a comment |
$begingroup$
Define
begin{align*}
f(x) = -xlog(x) - log(1-x)(log(x) + 1 - x)
end{align*}
It suffices to prove $f(x) ge 0$ for $x in [0, 1]$. Note that
begin{align*}
f(0) &= f(1) = 0 \
f'(x) &= frac{(x-1)log(1-x)}{x} + frac{xlog(x)}{1-x} begin{cases}
> 0 & x < frac{1}{2} \
= 0 & x = frac{1}{2} \
< 0 & x > frac{1}{2}
end{cases}
end{align*}
Therefore, $f(x) ge min(f(0), f(1)) = 0$ for $x in [0, 1]$, as desired.
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
$endgroup$
add a comment |
$begingroup$
Define
begin{align*}
f(x) = -xlog(x) - log(1-x)(log(x) + 1 - x)
end{align*}
It suffices to prove $f(x) ge 0$ for $x in [0, 1]$. Note that
begin{align*}
f(0) &= f(1) = 0 \
f'(x) &= frac{(x-1)log(1-x)}{x} + frac{xlog(x)}{1-x} begin{cases}
> 0 & x < frac{1}{2} \
= 0 & x = frac{1}{2} \
< 0 & x > frac{1}{2}
end{cases}
end{align*}
Therefore, $f(x) ge min(f(0), f(1)) = 0$ for $x in [0, 1]$, as desired.
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
$endgroup$
Define
begin{align*}
f(x) = -xlog(x) - log(1-x)(log(x) + 1 - x)
end{align*}
It suffices to prove $f(x) ge 0$ for $x in [0, 1]$. Note that
begin{align*}
f(0) &= f(1) = 0 \
f'(x) &= frac{(x-1)log(1-x)}{x} + frac{xlog(x)}{1-x} begin{cases}
> 0 & x < frac{1}{2} \
= 0 & x = frac{1}{2} \
< 0 & x > frac{1}{2}
end{cases}
end{align*}
Therefore, $f(x) ge min(f(0), f(1)) = 0$ for $x in [0, 1]$, as desired.
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
answered Jan 11 at 18:36
Tom ChenTom Chen
913513
913513
add a comment |
add a comment |
$begingroup$
Consider the function
$$f(x)=xln (x)+ ln(1-x),(ln (x) + 1 - x) $$ and notice the symmetry which makes that
$$f(x)+f(1-x)=0$$ Compute $f(x)$ anywhere to show that is negative. Using $x=frac 12$, you have $(log (2)-1) log (2)$ which is the minimum value.
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
$begingroup$
Consider the function
$$f(x)=xln (x)+ ln(1-x),(ln (x) + 1 - x) $$ and notice the symmetry which makes that
$$f(x)+f(1-x)=0$$ Compute $f(x)$ anywhere to show that is negative. Using $x=frac 12$, you have $(log (2)-1) log (2)$ which is the minimum value.
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
$begingroup$
Consider the function
$$f(x)=xln (x)+ ln(1-x),(ln (x) + 1 - x) $$ and notice the symmetry which makes that
$$f(x)+f(1-x)=0$$ Compute $f(x)$ anywhere to show that is negative. Using $x=frac 12$, you have $(log (2)-1) log (2)$ which is the minimum value.
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
Consider the function
$$f(x)=xln (x)+ ln(1-x),(ln (x) + 1 - x) $$ and notice the symmetry which makes that
$$f(x)+f(1-x)=0$$ Compute $f(x)$ anywhere to show that is negative. Using $x=frac 12$, you have $(log (2)-1) log (2)$ which is the minimum value.
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 7:04
Claude LeiboviciClaude Leibovici
121k1157133
121k1157133
add a comment |
add a comment |
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1
$begingroup$
prove/disprove, not "approve/disapprove"
$endgroup$
– rschwieb
Jan 11 at 17:49
$begingroup$
Do you mean $$-xln(x)le ln(1-x)(ln(x)+1-x)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:52
$begingroup$
@Dr.SonnhardGraubner yes I mean that
$endgroup$
– Weez Khan
Jan 11 at 17:53
$begingroup$
I think you mean $ge$, not $le$. The inequality you wrote down is not true for $0 < x < 1$, but if you switch it to $ge$ instead of $le$ then it is true.
$endgroup$
– 6005
Jan 11 at 17:55
$begingroup$
This inequalitiy is not true, take $$x=frac{1}{2}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 17:56