Prove an inequality concerning $ln$ function
$begingroup$
For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
$$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$
I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.
real-analysis derivatives inequality
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
asked Jan 6 at 3:51
zbh2047zbh2047
568
$endgroup$
add a comment |
$begingroup$
For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
$$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$
I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.
real-analysis derivatives inequality
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
asked Jan 6 at 3:51
zbh2047zbh2047
568
$endgroup$
add a comment |
$begingroup$
For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
$$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$
I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.
real-analysis derivatives inequality
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
asked Jan 6 at 3:51
zbh2047zbh2047
568
$endgroup$
For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
$$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$
I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.
real-analysis derivatives inequality
real-analysis derivatives inequality
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
asked Jan 6 at 3:51
zbh2047zbh2047
568
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
asked Jan 6 at 3:51
zbh2047zbh2047
568
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
edited Jan 6 at 5:55
Michael Rozenberg
108k1895200
108k1895200
asked Jan 6 at 3:51
zbh2047zbh2047
568
asked Jan 6 at 3:51
zbh2047zbh2047
568
asked Jan 6 at 3:51
zbh2047zbh2047
568
568
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The hint.
We need to prove that $f(delta)geq0,$ where
$$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
$$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
$$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
Can you end it now?
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Oh! Using Taylor expansion conclude the proof. Thanks!
$endgroup$
– zbh2047
Jan 6 at 8:05
$begingroup$
@zbh2047 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 6 at 8:55
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The hint.
We need to prove that $f(delta)geq0,$ where
$$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
$$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
$$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
Can you end it now?
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Oh! Using Taylor expansion conclude the proof. Thanks!
$endgroup$
– zbh2047
Jan 6 at 8:05
$begingroup$
@zbh2047 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 6 at 8:55
add a comment |
$begingroup$
The hint.
We need to prove that $f(delta)geq0,$ where
$$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
$$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
$$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
Can you end it now?
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Oh! Using Taylor expansion conclude the proof. Thanks!
$endgroup$
– zbh2047
Jan 6 at 8:05
$begingroup$
@zbh2047 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 6 at 8:55
add a comment |
$begingroup$
The hint.
We need to prove that $f(delta)geq0,$ where
$$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
$$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
$$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
Can you end it now?
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
The hint.
We need to prove that $f(delta)geq0,$ where
$$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
$$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
$$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
Can you end it now?
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 6 at 6:15
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 6 at 5:53
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
Oh! Using Taylor expansion conclude the proof. Thanks!
$endgroup$
– zbh2047
Jan 6 at 8:05
$begingroup$
@zbh2047 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 6 at 8:55
add a comment |
$begingroup$
Oh! Using Taylor expansion conclude the proof. Thanks!
$endgroup$
– zbh2047
Jan 6 at 8:05
$begingroup$
@zbh2047 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 6 at 8:55
$begingroup$
Oh! Using Taylor expansion conclude the proof. Thanks!
$endgroup$
– zbh2047
Jan 6 at 8:05
$begingroup$
Oh! Using Taylor expansion conclude the proof. Thanks!
$endgroup$
– zbh2047
Jan 6 at 8:05
$begingroup$
@zbh2047 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 6 at 8:55
$begingroup$
@zbh2047 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 6 at 8:55
add a comment |
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