Mixing
Minimum Normalizer
$begingroup$
Let $x_1ge x_2gedots ge x_n> 0$. What is the minimum value of $k$ such that $$frac{nkx_i}{x_1+dots+x_n}ge 1quad forall i.$$
I tried a bit hoping that $k=1+frac{1}{2}+dots+frac{1}{n}$ would work. But not getting any clue. Any hint is greatly appreciated.
calculus
asked Jan 13 at 12:06
KumaraKumara
221118
$endgroup$
add a comment |
$begingroup$
Let $x_1ge x_2gedots ge x_n> 0$. What is the minimum value of $k$ such that $$frac{nkx_i}{x_1+dots+x_n}ge 1quad forall i.$$
I tried a bit hoping that $k=1+frac{1}{2}+dots+frac{1}{n}$ would work. But not getting any clue. Any hint is greatly appreciated.
calculus
asked Jan 13 at 12:06
KumaraKumara
221118
$endgroup$
add a comment |
$begingroup$
Let $x_1ge x_2gedots ge x_n> 0$. What is the minimum value of $k$ such that $$frac{nkx_i}{x_1+dots+x_n}ge 1quad forall i.$$
I tried a bit hoping that $k=1+frac{1}{2}+dots+frac{1}{n}$ would work. But not getting any clue. Any hint is greatly appreciated.
calculus
asked Jan 13 at 12:06
KumaraKumara
221118
$endgroup$
Let $x_1ge x_2gedots ge x_n> 0$. What is the minimum value of $k$ such that $$frac{nkx_i}{x_1+dots+x_n}ge 1quad forall i.$$
I tried a bit hoping that $k=1+frac{1}{2}+dots+frac{1}{n}$ would work. But not getting any clue. Any hint is greatly appreciated.
calculus
calculus
asked Jan 13 at 12:06
KumaraKumara
221118
asked Jan 13 at 12:06
KumaraKumara
221118
asked Jan 13 at 12:06
KumaraKumara
221118
asked Jan 13 at 12:06
KumaraKumara
221118
asked Jan 13 at 12:06
KumaraKumara
221118
221118
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
You want to find the minimum $k>0$ such that
$$
k, x_igeq overline{x}, quad forall i = 1, ldots, n,
qquad
text{where} overline{x} := frac{x_1 + cdots + x_n}{n},.
$$
Since, by assumption, $x_1 geq x_2 geq cdots geq x_n > 0$, the above $n$ inequalities are satisfied if and only if $k, x_n geq overline{x}$. Hence, the minimum of such values of $k$ is $k = overline{x} / x_n$.
answered Jan 13 at 12:25
RigelRigel
11.2k11320
$endgroup$
$begingroup$
Sure, you are right, for some reason I was considering the reversed inequality!
$endgroup$
– Rigel
Jan 13 at 12:49
$begingroup$
Thanks. That's right. However, is it possible to obtain the bound in terms of $k=1+frac{1}{2}+dots+frac{1}{n}$ possibly with some multiplicative (constant) factor?
$endgroup$
– Kumara
Jan 13 at 15:46
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
oldest
votes
$begingroup$
You want to find the minimum $k>0$ such that
$$
k, x_igeq overline{x}, quad forall i = 1, ldots, n,
qquad
text{where} overline{x} := frac{x_1 + cdots + x_n}{n},.
$$
Since, by assumption, $x_1 geq x_2 geq cdots geq x_n > 0$, the above $n$ inequalities are satisfied if and only if $k, x_n geq overline{x}$. Hence, the minimum of such values of $k$ is $k = overline{x} / x_n$.
answered Jan 13 at 12:25
RigelRigel
11.2k11320
$endgroup$
$begingroup$
Sure, you are right, for some reason I was considering the reversed inequality!
$endgroup$
– Rigel
Jan 13 at 12:49
$begingroup$
Thanks. That's right. However, is it possible to obtain the bound in terms of $k=1+frac{1}{2}+dots+frac{1}{n}$ possibly with some multiplicative (constant) factor?
$endgroup$
– Kumara
Jan 13 at 15:46
add a comment |
$begingroup$
You want to find the minimum $k>0$ such that
$$
k, x_igeq overline{x}, quad forall i = 1, ldots, n,
qquad
text{where} overline{x} := frac{x_1 + cdots + x_n}{n},.
$$
Since, by assumption, $x_1 geq x_2 geq cdots geq x_n > 0$, the above $n$ inequalities are satisfied if and only if $k, x_n geq overline{x}$. Hence, the minimum of such values of $k$ is $k = overline{x} / x_n$.
answered Jan 13 at 12:25
RigelRigel
11.2k11320
$endgroup$
$begingroup$
Sure, you are right, for some reason I was considering the reversed inequality!
$endgroup$
– Rigel
Jan 13 at 12:49
$begingroup$
Thanks. That's right. However, is it possible to obtain the bound in terms of $k=1+frac{1}{2}+dots+frac{1}{n}$ possibly with some multiplicative (constant) factor?
$endgroup$
– Kumara
Jan 13 at 15:46
add a comment |
$begingroup$
You want to find the minimum $k>0$ such that
$$
k, x_igeq overline{x}, quad forall i = 1, ldots, n,
qquad
text{where} overline{x} := frac{x_1 + cdots + x_n}{n},.
$$
Since, by assumption, $x_1 geq x_2 geq cdots geq x_n > 0$, the above $n$ inequalities are satisfied if and only if $k, x_n geq overline{x}$. Hence, the minimum of such values of $k$ is $k = overline{x} / x_n$.
answered Jan 13 at 12:25
RigelRigel
11.2k11320
$endgroup$
You want to find the minimum $k>0$ such that
$$
k, x_igeq overline{x}, quad forall i = 1, ldots, n,
qquad
text{where} overline{x} := frac{x_1 + cdots + x_n}{n},.
$$
Since, by assumption, $x_1 geq x_2 geq cdots geq x_n > 0$, the above $n$ inequalities are satisfied if and only if $k, x_n geq overline{x}$. Hence, the minimum of such values of $k$ is $k = overline{x} / x_n$.
answered Jan 13 at 12:25
RigelRigel
11.2k11320
edited Jan 13 at 12:48
answered Jan 13 at 12:25
RigelRigel
11.2k11320
answered Jan 13 at 12:25
RigelRigel
11.2k11320
answered Jan 13 at 12:25
RigelRigel
11.2k11320
11.2k11320
$begingroup$
Sure, you are right, for some reason I was considering the reversed inequality!
$endgroup$
– Rigel
Jan 13 at 12:49
$begingroup$
Thanks. That's right. However, is it possible to obtain the bound in terms of $k=1+frac{1}{2}+dots+frac{1}{n}$ possibly with some multiplicative (constant) factor?
$endgroup$
– Kumara
Jan 13 at 15:46
add a comment |
$begingroup$
Sure, you are right, for some reason I was considering the reversed inequality!
$endgroup$
– Rigel
Jan 13 at 12:49
$begingroup$
Thanks. That's right. However, is it possible to obtain the bound in terms of $k=1+frac{1}{2}+dots+frac{1}{n}$ possibly with some multiplicative (constant) factor?
$endgroup$
– Kumara
Jan 13 at 15:46
$begingroup$
Sure, you are right, for some reason I was considering the reversed inequality!
$endgroup$
– Rigel
Jan 13 at 12:49
$begingroup$
Sure, you are right, for some reason I was considering the reversed inequality!
$endgroup$
– Rigel
Jan 13 at 12:49
$begingroup$
Thanks. That's right. However, is it possible to obtain the bound in terms of $k=1+frac{1}{2}+dots+frac{1}{n}$ possibly with some multiplicative (constant) factor?
$endgroup$
– Kumara
Jan 13 at 15:46
$begingroup$
Thanks. That's right. However, is it possible to obtain the bound in terms of $k=1+frac{1}{2}+dots+frac{1}{n}$ possibly with some multiplicative (constant) factor?
$endgroup$
– Kumara
Jan 13 at 15:46
add a comment |
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