Lower bound for the function $h(x)=(1+1/x)log(1+x)-1$ for $x > 0$.
$begingroup$
I am reading a paper and in the paper, there is a result having no proof. I was wondering if anyone could give me a hint on this result. The result is the following:
$$
h(x)geq begin{cases}
x/4, & text{if } xleq4;\
(1/2)log x, & text{if } xgeq4,
end{cases}
$$
where the function $h$ is defined for all $x > 0$ by
$$h(x)=(1+1/x)log(1+x)-1.$$
Thanks in advance.
real-analysis inequality
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
$endgroup$
add a comment |
$begingroup$
I am reading a paper and in the paper, there is a result having no proof. I was wondering if anyone could give me a hint on this result. The result is the following:
$$
h(x)geq begin{cases}
x/4, & text{if } xleq4;\
(1/2)log x, & text{if } xgeq4,
end{cases}
$$
where the function $h$ is defined for all $x > 0$ by
$$h(x)=(1+1/x)log(1+x)-1.$$
Thanks in advance.
real-analysis inequality
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
$endgroup$
$begingroup$
Presumably you are requiring $x > 0$, as it's not true if $-1 < x < 0$.
$endgroup$
– Robert Israel
Dec 16 '18 at 3:53
$begingroup$
Yes, exactly. x is positive.
$endgroup$
– M.Shen
Dec 16 '18 at 4:00
add a comment |
$begingroup$
I am reading a paper and in the paper, there is a result having no proof. I was wondering if anyone could give me a hint on this result. The result is the following:
$$
h(x)geq begin{cases}
x/4, & text{if } xleq4;\
(1/2)log x, & text{if } xgeq4,
end{cases}
$$
where the function $h$ is defined for all $x > 0$ by
$$h(x)=(1+1/x)log(1+x)-1.$$
Thanks in advance.
real-analysis inequality
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
$endgroup$
I am reading a paper and in the paper, there is a result having no proof. I was wondering if anyone could give me a hint on this result. The result is the following:
$$
h(x)geq begin{cases}
x/4, & text{if } xleq4;\
(1/2)log x, & text{if } xgeq4,
end{cases}
$$
where the function $h$ is defined for all $x > 0$ by
$$h(x)=(1+1/x)log(1+x)-1.$$
Thanks in advance.
real-analysis inequality
real-analysis inequality
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
edited Dec 16 '18 at 5:46
Brahadeesh
6,46942363
6,46942363
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
asked Dec 16 '18 at 3:23
M.ShenM.Shen
284
284
$begingroup$
Presumably you are requiring $x > 0$, as it's not true if $-1 < x < 0$.
$endgroup$
– Robert Israel
Dec 16 '18 at 3:53
$begingroup$
Yes, exactly. x is positive.
$endgroup$
– M.Shen
Dec 16 '18 at 4:00
add a comment |
$begingroup$
Presumably you are requiring $x > 0$, as it's not true if $-1 < x < 0$.
$endgroup$
– Robert Israel
Dec 16 '18 at 3:53
$begingroup$
Yes, exactly. x is positive.
$endgroup$
– M.Shen
Dec 16 '18 at 4:00
$begingroup$
Presumably you are requiring $x > 0$, as it's not true if $-1 < x < 0$.
$endgroup$
– Robert Israel
Dec 16 '18 at 3:53
$begingroup$
Presumably you are requiring $x > 0$, as it's not true if $-1 < x < 0$.
$endgroup$
– Robert Israel
Dec 16 '18 at 3:53
$begingroup$
Yes, exactly. x is positive.
$endgroup$
– M.Shen
Dec 16 '18 at 4:00
$begingroup$
Yes, exactly. x is positive.
$endgroup$
– M.Shen
Dec 16 '18 at 4:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $0<xleq 4$ we need to prove that $$left(1+frac{1}{x}right)ln(1+x)-1geqfrac{x}{4}$$ or $f(x)geq0,$ where
$$f(x)=ln(1+x)-frac{4x+x^2}{4(1+x)}.$$
But, $$f'(x)=frac{x(2-x)}{3(1+x)^2},$$
which says that $$min_{0leq xleq4}f=min{f(0),f(4)}=f(0)=0.$$
By the same way you can prove the second inequality:
after calculating of derivative use AM-GM.
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
For $0<xleq 4$ we need to prove that $$left(1+frac{1}{x}right)ln(1+x)-1geqfrac{x}{4}$$ or $f(x)geq0,$ where
$$f(x)=ln(1+x)-frac{4x+x^2}{4(1+x)}.$$
But, $$f'(x)=frac{x(2-x)}{3(1+x)^2},$$
which says that $$min_{0leq xleq4}f=min{f(0),f(4)}=f(0)=0.$$
By the same way you can prove the second inequality:
after calculating of derivative use AM-GM.
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
$begingroup$
For $0<xleq 4$ we need to prove that $$left(1+frac{1}{x}right)ln(1+x)-1geqfrac{x}{4}$$ or $f(x)geq0,$ where
$$f(x)=ln(1+x)-frac{4x+x^2}{4(1+x)}.$$
But, $$f'(x)=frac{x(2-x)}{3(1+x)^2},$$
which says that $$min_{0leq xleq4}f=min{f(0),f(4)}=f(0)=0.$$
By the same way you can prove the second inequality:
after calculating of derivative use AM-GM.
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
$begingroup$
For $0<xleq 4$ we need to prove that $$left(1+frac{1}{x}right)ln(1+x)-1geqfrac{x}{4}$$ or $f(x)geq0,$ where
$$f(x)=ln(1+x)-frac{4x+x^2}{4(1+x)}.$$
But, $$f'(x)=frac{x(2-x)}{3(1+x)^2},$$
which says that $$min_{0leq xleq4}f=min{f(0),f(4)}=f(0)=0.$$
By the same way you can prove the second inequality:
after calculating of derivative use AM-GM.
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
For $0<xleq 4$ we need to prove that $$left(1+frac{1}{x}right)ln(1+x)-1geqfrac{x}{4}$$ or $f(x)geq0,$ where
$$f(x)=ln(1+x)-frac{4x+x^2}{4(1+x)}.$$
But, $$f'(x)=frac{x(2-x)}{3(1+x)^2},$$
which says that $$min_{0leq xleq4}f=min{f(0),f(4)}=f(0)=0.$$
By the same way you can prove the second inequality:
after calculating of derivative use AM-GM.
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
edited Dec 16 '18 at 6:19
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
answered Dec 16 '18 at 6:09
Michael RozenbergMichael Rozenberg
104k1892197
104k1892197
add a comment |
add a comment |
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$begingroup$
Presumably you are requiring $x > 0$, as it's not true if $-1 < x < 0$.
$endgroup$
– Robert Israel
Dec 16 '18 at 3:53
$begingroup$
Yes, exactly. x is positive.
$endgroup$
– M.Shen
Dec 16 '18 at 4:00