prove $ H(x)geq ln(x)(ln (1 - x))$ for $ 0lt xlt 1$
$begingroup$
The equation below is true i have proved it using values between 0 and 1 but how can I prove it mathematically? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ H(x)geq ln(x)(ln (1 - x)) qquadtext{for}qquad 0lt xlt 1$$
where $$ H(x) = -x ln(x) - (1-x)ln(1-x)$$
Any help will be quite useful to me. Thanks.
inequality logarithms
edited Jan 13 at 14:09
Martin R
27.9k33255
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
$endgroup$
add a comment |
$begingroup$
The equation below is true i have proved it using values between 0 and 1 but how can I prove it mathematically? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ H(x)geq ln(x)(ln (1 - x)) qquadtext{for}qquad 0lt xlt 1$$
where $$ H(x) = -x ln(x) - (1-x)ln(1-x)$$
Any help will be quite useful to me. Thanks.
inequality logarithms
edited Jan 13 at 14:09
Martin R
27.9k33255
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
$endgroup$
add a comment |
$begingroup$
The equation below is true i have proved it using values between 0 and 1 but how can I prove it mathematically? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ H(x)geq ln(x)(ln (1 - x)) qquadtext{for}qquad 0lt xlt 1$$
where $$ H(x) = -x ln(x) - (1-x)ln(1-x)$$
Any help will be quite useful to me. Thanks.
inequality logarithms
edited Jan 13 at 14:09
Martin R
27.9k33255
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
$endgroup$
The equation below is true i have proved it using values between 0 and 1 but how can I prove it mathematically? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ H(x)geq ln(x)(ln (1 - x)) qquadtext{for}qquad 0lt xlt 1$$
where $$ H(x) = -x ln(x) - (1-x)ln(1-x)$$
Any help will be quite useful to me. Thanks.
inequality logarithms
inequality logarithms
edited Jan 13 at 14:09
Martin R
27.9k33255
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
edited Jan 13 at 14:09
Martin R
27.9k33255
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
edited Jan 13 at 14:09
Martin R
27.9k33255
edited Jan 13 at 14:09
Martin R
27.9k33255
edited Jan 13 at 14:09
Martin R
27.9k33255
27.9k33255
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
asked Jan 12 at 11:00
Weez KhanWeez Khan
135
135
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We have to show that
$$
f(x) = x ln(x) + (1-x) ln(1-x) + ln(x) ln(1-x) \
= bigl( ln(x) + 1 - x bigr) bigl( ln(1-x) + x bigr) - x(1-x)
$$
is $le 0$ for $0 < x < 1$. Using the “well-known” estimates
$$
frac{x-1}{x} le ln(x) le x-1 quad (text{for } x > 0)
$$
(see for example How can I prove that $ frac {x-1}{x}leq log xleq x-1$), the first factor can be estimated as
$$
0 ge ln(x) + 1 - x ge frac{x-1}{x} + 1-x = -frac{(1-x)^2}{x}
$$
and the second factor as
$$
0 ge ln(1-x) + x ge frac{-x}{1-x} + x = -frac{x^2}{1-x} , .
$$
It follows that
$$
f(x) le frac{(1-x)^2}{x} cdot frac{x^2}{1-x} - x(1-x) = 0 , .
$$
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
$begingroup$
Note that for $0lt xlt1$,
$$
x=int_0^x1,mathrm{d}tleoverbrace{int_0^xfrac1{1-t},mathrm{d}t}^{-log(1-x)}leint_0^xfrac1{1-x},mathrm{d}t=frac{x}{1-x}tag1
$$
Negating gives,
$$
-frac{x}{1-x}lelog(1-x)le-xtag2
$$
Therefore,
$$
-frac{x^2}{1-x}le x+log(1-x)le0tag3
$$
and substituting $xmapsto1-x$,
$$
-frac{(1-x)^2}{x}le(1-x)+log(x)le0tag4
$$
Multiplying inequalities $(3)$ and $(4)$ gives
$$
[x+log(1-x)][(1-x)+log(x)]le x(1-x)tag5
$$
which upon rearrangement yields
$$
log(1-x)log(x)le-xlog(x)-(1-x)log(x)tag6
$$
which is the inequality sought.
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
$endgroup$
add a comment |
$begingroup$
One approach is to show that the function $$G(x) = ln(x) ln(1-x) - H(x)$$ is convex, i.e. $G''(x) geq 0$ for $0 < x < 1$. Since $G$ is continuous on $[0,1]$ and $G(0) = G(1) = 0$, it follows that $G(x) leq 0$ on $[0,1]$ yielding the desired inequality.
A direct calculation gives
$$
G''(x) = frac{-ln(x)}{(1-x)^2} + frac{-ln(1-x)}{x^2} - frac{1}{x(1-x)}.
$$
The first two terms have a "good" sign, but the third term has a "bad" sign for $0 < x < 1$. Using the elementary inequalities $-ln(1-x) geq x, -ln(x) geq 1-x$ for $0 leq x leq 1$, we get
$
G''(x) geq 0.
$
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have to show that
$$
f(x) = x ln(x) + (1-x) ln(1-x) + ln(x) ln(1-x) \
= bigl( ln(x) + 1 - x bigr) bigl( ln(1-x) + x bigr) - x(1-x)
$$
is $le 0$ for $0 < x < 1$. Using the “well-known” estimates
$$
frac{x-1}{x} le ln(x) le x-1 quad (text{for } x > 0)
$$
(see for example How can I prove that $ frac {x-1}{x}leq log xleq x-1$), the first factor can be estimated as
$$
0 ge ln(x) + 1 - x ge frac{x-1}{x} + 1-x = -frac{(1-x)^2}{x}
$$
and the second factor as
$$
0 ge ln(1-x) + x ge frac{-x}{1-x} + x = -frac{x^2}{1-x} , .
$$
It follows that
$$
f(x) le frac{(1-x)^2}{x} cdot frac{x^2}{1-x} - x(1-x) = 0 , .
$$
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
$begingroup$
We have to show that
$$
f(x) = x ln(x) + (1-x) ln(1-x) + ln(x) ln(1-x) \
= bigl( ln(x) + 1 - x bigr) bigl( ln(1-x) + x bigr) - x(1-x)
$$
is $le 0$ for $0 < x < 1$. Using the “well-known” estimates
$$
frac{x-1}{x} le ln(x) le x-1 quad (text{for } x > 0)
$$
(see for example How can I prove that $ frac {x-1}{x}leq log xleq x-1$), the first factor can be estimated as
$$
0 ge ln(x) + 1 - x ge frac{x-1}{x} + 1-x = -frac{(1-x)^2}{x}
$$
and the second factor as
$$
0 ge ln(1-x) + x ge frac{-x}{1-x} + x = -frac{x^2}{1-x} , .
$$
It follows that
$$
f(x) le frac{(1-x)^2}{x} cdot frac{x^2}{1-x} - x(1-x) = 0 , .
$$
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
$begingroup$
We have to show that
$$
f(x) = x ln(x) + (1-x) ln(1-x) + ln(x) ln(1-x) \
= bigl( ln(x) + 1 - x bigr) bigl( ln(1-x) + x bigr) - x(1-x)
$$
is $le 0$ for $0 < x < 1$. Using the “well-known” estimates
$$
frac{x-1}{x} le ln(x) le x-1 quad (text{for } x > 0)
$$
(see for example How can I prove that $ frac {x-1}{x}leq log xleq x-1$), the first factor can be estimated as
$$
0 ge ln(x) + 1 - x ge frac{x-1}{x} + 1-x = -frac{(1-x)^2}{x}
$$
and the second factor as
$$
0 ge ln(1-x) + x ge frac{-x}{1-x} + x = -frac{x^2}{1-x} , .
$$
It follows that
$$
f(x) le frac{(1-x)^2}{x} cdot frac{x^2}{1-x} - x(1-x) = 0 , .
$$
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
$endgroup$
We have to show that
$$
f(x) = x ln(x) + (1-x) ln(1-x) + ln(x) ln(1-x) \
= bigl( ln(x) + 1 - x bigr) bigl( ln(1-x) + x bigr) - x(1-x)
$$
is $le 0$ for $0 < x < 1$. Using the “well-known” estimates
$$
frac{x-1}{x} le ln(x) le x-1 quad (text{for } x > 0)
$$
(see for example How can I prove that $ frac {x-1}{x}leq log xleq x-1$), the first factor can be estimated as
$$
0 ge ln(x) + 1 - x ge frac{x-1}{x} + 1-x = -frac{(1-x)^2}{x}
$$
and the second factor as
$$
0 ge ln(1-x) + x ge frac{-x}{1-x} + x = -frac{x^2}{1-x} , .
$$
It follows that
$$
f(x) le frac{(1-x)^2}{x} cdot frac{x^2}{1-x} - x(1-x) = 0 , .
$$
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
edited Jan 13 at 17:35
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
answered Jan 12 at 18:37
Martin RMartin R
27.9k33255
27.9k33255
add a comment |
add a comment |
$begingroup$
Note that for $0lt xlt1$,
$$
x=int_0^x1,mathrm{d}tleoverbrace{int_0^xfrac1{1-t},mathrm{d}t}^{-log(1-x)}leint_0^xfrac1{1-x},mathrm{d}t=frac{x}{1-x}tag1
$$
Negating gives,
$$
-frac{x}{1-x}lelog(1-x)le-xtag2
$$
Therefore,
$$
-frac{x^2}{1-x}le x+log(1-x)le0tag3
$$
and substituting $xmapsto1-x$,
$$
-frac{(1-x)^2}{x}le(1-x)+log(x)le0tag4
$$
Multiplying inequalities $(3)$ and $(4)$ gives
$$
[x+log(1-x)][(1-x)+log(x)]le x(1-x)tag5
$$
which upon rearrangement yields
$$
log(1-x)log(x)le-xlog(x)-(1-x)log(x)tag6
$$
which is the inequality sought.
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
$endgroup$
add a comment |
$begingroup$
Note that for $0lt xlt1$,
$$
x=int_0^x1,mathrm{d}tleoverbrace{int_0^xfrac1{1-t},mathrm{d}t}^{-log(1-x)}leint_0^xfrac1{1-x},mathrm{d}t=frac{x}{1-x}tag1
$$
Negating gives,
$$
-frac{x}{1-x}lelog(1-x)le-xtag2
$$
Therefore,
$$
-frac{x^2}{1-x}le x+log(1-x)le0tag3
$$
and substituting $xmapsto1-x$,
$$
-frac{(1-x)^2}{x}le(1-x)+log(x)le0tag4
$$
Multiplying inequalities $(3)$ and $(4)$ gives
$$
[x+log(1-x)][(1-x)+log(x)]le x(1-x)tag5
$$
which upon rearrangement yields
$$
log(1-x)log(x)le-xlog(x)-(1-x)log(x)tag6
$$
which is the inequality sought.
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
$endgroup$
add a comment |
$begingroup$
Note that for $0lt xlt1$,
$$
x=int_0^x1,mathrm{d}tleoverbrace{int_0^xfrac1{1-t},mathrm{d}t}^{-log(1-x)}leint_0^xfrac1{1-x},mathrm{d}t=frac{x}{1-x}tag1
$$
Negating gives,
$$
-frac{x}{1-x}lelog(1-x)le-xtag2
$$
Therefore,
$$
-frac{x^2}{1-x}le x+log(1-x)le0tag3
$$
and substituting $xmapsto1-x$,
$$
-frac{(1-x)^2}{x}le(1-x)+log(x)le0tag4
$$
Multiplying inequalities $(3)$ and $(4)$ gives
$$
[x+log(1-x)][(1-x)+log(x)]le x(1-x)tag5
$$
which upon rearrangement yields
$$
log(1-x)log(x)le-xlog(x)-(1-x)log(x)tag6
$$
which is the inequality sought.
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
$endgroup$
Note that for $0lt xlt1$,
$$
x=int_0^x1,mathrm{d}tleoverbrace{int_0^xfrac1{1-t},mathrm{d}t}^{-log(1-x)}leint_0^xfrac1{1-x},mathrm{d}t=frac{x}{1-x}tag1
$$
Negating gives,
$$
-frac{x}{1-x}lelog(1-x)le-xtag2
$$
Therefore,
$$
-frac{x^2}{1-x}le x+log(1-x)le0tag3
$$
and substituting $xmapsto1-x$,
$$
-frac{(1-x)^2}{x}le(1-x)+log(x)le0tag4
$$
Multiplying inequalities $(3)$ and $(4)$ gives
$$
[x+log(1-x)][(1-x)+log(x)]le x(1-x)tag5
$$
which upon rearrangement yields
$$
log(1-x)log(x)le-xlog(x)-(1-x)log(x)tag6
$$
which is the inequality sought.
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
answered Jan 13 at 13:35
robjohn♦robjohn
266k27306630
266k27306630
add a comment |
add a comment |
$begingroup$
One approach is to show that the function $$G(x) = ln(x) ln(1-x) - H(x)$$ is convex, i.e. $G''(x) geq 0$ for $0 < x < 1$. Since $G$ is continuous on $[0,1]$ and $G(0) = G(1) = 0$, it follows that $G(x) leq 0$ on $[0,1]$ yielding the desired inequality.
A direct calculation gives
$$
G''(x) = frac{-ln(x)}{(1-x)^2} + frac{-ln(1-x)}{x^2} - frac{1}{x(1-x)}.
$$
The first two terms have a "good" sign, but the third term has a "bad" sign for $0 < x < 1$. Using the elementary inequalities $-ln(1-x) geq x, -ln(x) geq 1-x$ for $0 leq x leq 1$, we get
$
G''(x) geq 0.
$
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
$endgroup$
add a comment |
$begingroup$
One approach is to show that the function $$G(x) = ln(x) ln(1-x) - H(x)$$ is convex, i.e. $G''(x) geq 0$ for $0 < x < 1$. Since $G$ is continuous on $[0,1]$ and $G(0) = G(1) = 0$, it follows that $G(x) leq 0$ on $[0,1]$ yielding the desired inequality.
A direct calculation gives
$$
G''(x) = frac{-ln(x)}{(1-x)^2} + frac{-ln(1-x)}{x^2} - frac{1}{x(1-x)}.
$$
The first two terms have a "good" sign, but the third term has a "bad" sign for $0 < x < 1$. Using the elementary inequalities $-ln(1-x) geq x, -ln(x) geq 1-x$ for $0 leq x leq 1$, we get
$
G''(x) geq 0.
$
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
$endgroup$
add a comment |
$begingroup$
One approach is to show that the function $$G(x) = ln(x) ln(1-x) - H(x)$$ is convex, i.e. $G''(x) geq 0$ for $0 < x < 1$. Since $G$ is continuous on $[0,1]$ and $G(0) = G(1) = 0$, it follows that $G(x) leq 0$ on $[0,1]$ yielding the desired inequality.
A direct calculation gives
$$
G''(x) = frac{-ln(x)}{(1-x)^2} + frac{-ln(1-x)}{x^2} - frac{1}{x(1-x)}.
$$
The first two terms have a "good" sign, but the third term has a "bad" sign for $0 < x < 1$. Using the elementary inequalities $-ln(1-x) geq x, -ln(x) geq 1-x$ for $0 leq x leq 1$, we get
$
G''(x) geq 0.
$
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
$endgroup$
One approach is to show that the function $$G(x) = ln(x) ln(1-x) - H(x)$$ is convex, i.e. $G''(x) geq 0$ for $0 < x < 1$. Since $G$ is continuous on $[0,1]$ and $G(0) = G(1) = 0$, it follows that $G(x) leq 0$ on $[0,1]$ yielding the desired inequality.
A direct calculation gives
$$
G''(x) = frac{-ln(x)}{(1-x)^2} + frac{-ln(1-x)}{x^2} - frac{1}{x(1-x)}.
$$
The first two terms have a "good" sign, but the third term has a "bad" sign for $0 < x < 1$. Using the elementary inequalities $-ln(1-x) geq x, -ln(x) geq 1-x$ for $0 leq x leq 1$, we get
$
G''(x) geq 0.
$
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
answered Jan 12 at 13:05
Shankar VenkataramaniShankar Venkataramani
11
11
add a comment |
add a comment |
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