Mixing
Proving $b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$ where $0 leq b leq a leq 1$
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I am trying to prove an inequality which boils down to the following:
$$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$
Another equivalent formulation is:
$$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$
How can I prove it?
inequality logarithms exponentiation
asked Jan 29 at 4:51
hi15hi15
1586
$endgroup$
add a comment |
$begingroup$
I am trying to prove an inequality which boils down to the following:
$$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$
Another equivalent formulation is:
$$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$
How can I prove it?
inequality logarithms exponentiation
asked Jan 29 at 4:51
hi15hi15
1586
$endgroup$
add a comment |
$begingroup$
I am trying to prove an inequality which boils down to the following:
$$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$
Another equivalent formulation is:
$$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$
How can I prove it?
inequality logarithms exponentiation
asked Jan 29 at 4:51
hi15hi15
1586
$endgroup$
I am trying to prove an inequality which boils down to the following:
$$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$
Another equivalent formulation is:
$$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$
How can I prove it?
inequality logarithms exponentiation
inequality logarithms exponentiation
asked Jan 29 at 4:51
hi15hi15
1586
asked Jan 29 at 4:51
hi15hi15
1586
asked Jan 29 at 4:51
hi15hi15
1586
asked Jan 29 at 4:51
hi15hi15
1586
asked Jan 29 at 4:51
hi15hi15
1586
1586
add a comment |
add a comment |
1 Answer
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$begingroup$
For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
So $f_a$ is maximum at $x=a$.
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
So $f_a$ is maximum at $x=a$.
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
$begingroup$
For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
So $f_a$ is maximum at $x=a$.
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
$begingroup$
For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
So $f_a$ is maximum at $x=a$.
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
$endgroup$
For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
So $f_a$ is maximum at $x=a$.
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 5:00
Stefan LafonStefan Lafon
3,005212
3,005212
add a comment |
add a comment |
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